Exercises — Solutions

1. 1.13 marksNCERT Exercises, Chapter 1

   Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

   Hint. Classify by the physical state of the solution itself, then by the state of the solute within it.

   A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within limits. The component present in the larger amount is the solvent and the others are solutes.

   Solutions are classified by the state of the solution, and within each by the state of the solute, giving nine types in all:

   Solution state	Solute state	Example
   Gaseous	Gas	Mixture of oxygen and nitrogen
   Gaseous	Liquid	Chloroform vapour in nitrogen
   Gaseous	Solid	Camphor in nitrogen gas
   Liquid	Gas	Oxygen dissolved in water
   Liquid	Liquid	Ethanol dissolved in water
   Liquid	Solid	Glucose dissolved in water
   Solid	Gas	Hydrogen in palladium
   Solid	Liquid	Amalgam of mercury with sodium
   Solid	Solid	Copper dissolved in gold

   A binary solution contains just two components, and that is the case this chapter deals with throughout, because two-component systems are where Raoult's law and the colligative properties can be stated most simply.

   ✦ A solution is a homogeneous mixture of two or more substances; there are nine types, classified by the physical state of the solution and of the solute

2. 1.21 markNCERT Exercises, Chapter 1

   Give an example of a solid solution in which the solute is a gas.

   Hint. Look for a gas absorbed into a metal lattice.

   A solid solution in which the solute is a gas is a solution of hydrogen in palladium.

   Palladium metal absorbs large volumes of hydrogen gas into the spaces of its crystal lattice, forming a genuinely homogeneous solid phase rather than a compound. This property is exploited industrially to purify hydrogen, since only hydrogen passes through the metal.

   Another acceptable example is a solution of hydrogen in platinum, which behaves in the same way.

   ✦ A solution of hydrogen gas in palladium metal

3. 1.34 marksNCERT Exercises, Chapter 1

   Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

   Hint. For each, be precise about whether the denominator refers to the solution or to the solvent alone.

   (i) Mole fraction is the ratio of the number of moles of one component to the total number of moles of all components in the solution:

   $x_A=\frac{n_A}{n_A+n_B}$

   It is dimensionless, and the mole fractions of all components sum to one.

   (ii) Molality is the number of moles of solute dissolved per kilogram of solvent:

   $m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

   Its unit is mol kg${}^{-1}$. Because it involves only masses, molality is independent of temperature.

   (iii) Molarity is the number of moles of solute per litre of solution:

   $M=\frac{\text{moles of solute}}{\text{volume of solution in L}}$

   Its unit is mol L${}^{-1}$. Since volume expands on heating, molarity does depend on temperature.

   (iv) Mass percentage is the mass of a component divided by the total mass of the solution, multiplied by 100.

   The distinction that matters most is between molality and molarity: one is per kilogram of solvent, the other per litre of solution, and only the first is temperature-independent.

   ✦ Mole fraction is moles of a component over total moles; molality is moles of solute per kg of solvent; molarity is moles of solute per litre of solution; mass percentage is mass of component over total mass times 100

4. 1.43 marksNCERT Exercises, Chapter 1

   Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL${}^{-1}$?

   Hint. Take 100 g of solution so the percentage becomes a mass directly, then use the density to get the volume.

   Take 100 g of solution, which then contains 68 g of HNO${}_3$. With $M({\text{HNO}}_3)=63$ g mol${}^{-1}$:

   $n=\frac{68}{63}=1.079\text{mol}$

   The density converts that 100 g into a volume:

   $V=\frac{100}{1.504}=66.49\text{mL}=0.06649\text{L}$

   Therefore:

   $\text{Molarity}=\frac{1.079}{0.06649}=16.23\text{M}$

   The value is high because the acid is both concentrated and dense, which is why concentrated nitric acid is always diluted before use.

   ✦ 16.23 M

5. 1.54 marksNCERT Exercises, Chapter 1

   A solution of glucose in water is labelled as 10% w/w. What would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL${}^{-1}$, then what shall be the molarity of the solution?

   Hint. Take 100 g of solution; the density is needed only for the molarity, not for the molality or mole fractions.

   Take 100 g of solution, giving 10 g of glucose and 90 g of water. With $M(\text{glucose})=180$ g mol${}^{-1}$:

   $n_{\text{glucose}}=\frac{10}{180}=0.0556\text{mol},n_{{\text{H}}_2\text{O}}=\frac{90}{18}=5.0\text{mol}$

   Molality uses the mass of water alone:

   $m=\frac{0.0556}{0.090}=0.617{\text{mol kg}}^{-1}$

   Mole fractions:

   $x_{\text{glucose}}=\frac{0.0556}{0.0556+5.0}=0.011,x_{{\text{H}}_2\text{O}}=0.989$

   Molarity needs the volume, so the density is used here and only here:

   $V=\frac{100}{1.2}=83.33\text{mL}=0.08333\text{L}$

   $M=\frac{0.0556}{0.08333}=0.667{\text{mol L}}^{-1}$

   ✦ Molality 0.617 mol/kg; x(glucose) = 0.011 and x(water) = 0.989; molarity 0.667 M

6. 1.64 marksNCERT Exercises, Chapter 1

   How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na${}_2$CO${}_3$ and NaHCO${}_3$ containing equimolar amounts of both?

   Hint. Write both neutralisation equations first, since the two carbonates consume different amounts of acid per mole.

   The two reactions consume different amounts of acid per mole, and that is the point of the question, since the carbonate is dibasic while the bicarbonate is monobasic:

   ${\text{Na}}_2{\text{CO}}_3+2\text{HCl}\to 2\text{NaCl}+{\text{H}}_2\text{O}+{\text{CO}}_2$

   ${\text{NaHCO}}_3+\text{HCl}\to \text{NaCl}+{\text{H}}_2\text{O}+{\text{CO}}_2$

   Let the mixture contain $n$ mol of each. With $M({\text{Na}}_2{\text{CO}}_3)=106$ and $M({\text{NaHCO}}_3)=84$ g mol${}^{-1}$:

   $106n+84n=1⟹190n=1⟹n=5.263\times 10^{-3}\text{mol}$

   The carbonate needs $2n$ mol of HCl and the bicarbonate $n$ mol, so:

   $n_{\text{HCl}}=3n=1.579\times 10^{-2}\text{mol}$

   $V=\frac{1.579\times 10^{-2}}{0.1}=0.1579\text{L}=157.9\text{mL}$

   ✦ 157.9 mL of 0.1 M HCl

7. 1.72 marksNCERT Exercises, Chapter 1

   A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

   Hint. Add the solute masses and add the total masses separately, then divide.

   Find the mass of solute contributed by each solution:

   $300\times 0.25=75\text{g},400\times 0.40=160\text{g}$

   So the mixture contains $75+160=235$ g of solute in a total mass of $300+400=700$ g.

   $\%\text{ solute}=\frac{235}{700}\times 100=33.57\%$

   and the solvent is therefore $66.43\%$.

   Note that the answer is not the average of 25 and 40, because the two solutions were mixed in unequal masses, so the more plentiful 40% solution pulls the result above the midpoint.

   ✦ 33.57% solute and 66.43% solvent

8. 1.84 marksNCERT Exercises, Chapter 1

   An antifreeze solution is prepared from 222.6 g of ethylene glycol (C${}_2$H${}_6$O${}_2$) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL${}^{-1}$, then what shall be the molarity of the solution?

   Hint. Molality uses the 200 g of water; molarity needs the total mass converted to a volume by the density.

   With $M({\text{C}}_2{\text{H}}_6{\text{O}}_2)=62$ g mol${}^{-1}$:

   $n=\frac{222.6}{62}=3.590\text{mol}$

   Molality is per kilogram of solvent, which is the 200 g of water:

   $m=\frac{3.590}{0.200}=17.95{\text{mol kg}}^{-1}$

   Molarity needs the volume of the whole solution. The total mass is $222.6+200=422.6$ g, so:

   $V=\frac{422.6}{1.072}=394.2\text{mL}=0.3942\text{L}$

   $M=\frac{3.590}{0.3942}=9.11{\text{mol L}}^{-1}$

   The two values differ by nearly a factor of two here, because the solution is extremely concentrated. In dilute aqueous solutions they nearly coincide, which is why the distinction is easy to forget.

   ✦ Molality 17.95 mol/kg; molarity 9.11 M

9. 1.94 marksNCERT Exercises, Chapter 1

   A sample of drinking water was found to be severely contaminated with chloroform (CHCl${}_3$) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample.

   Hint. Parts per million means parts per 10^6 by mass; for the molality, note that the solution is so dilute it is essentially pure water.

   (i) Parts per million means parts per $10^6$ by mass, so 15 ppm is 15 g of chloroform in $10^6$ g of solution:

   $\%\text{ by mass}=\frac{15}{10^6}\times 100=1.5\times 10^{-3}\%$

   (ii) For the molality, take that same $10^6$ g of solution. Since the contamination is so slight, the mass of water is essentially $10^6$ g, that is 1000 kg.

   With $M({\text{CHCl}}_3)=12+1+3(35.5)=119.5$ g mol${}^{-1}$:

   $n=\frac{15}{119.5}=0.1255\text{mol}$

   $m=\frac{0.1255}{1000}=1.255\times 10^{-4}{\text{mol kg}}^{-1}$

   ✦ (i) 1.5 x 10^-3 % by mass (ii) 1.255 x 10^-4 mol/kg

10. 1.103 marksNCERT Exercises, Chapter 1

    What role does the molecular interaction play in a solution of alcohol and water?

    Hint. Compare the strength of the hydrogen bonds present before mixing with those present afterwards.

    Both alcohol and water are strongly hydrogen bonded in the pure state. When they are mixed, the new alcohol-water hydrogen bonds that form are weaker than the original alcohol-alcohol and water-water bonds they replace.

    Three consequences follow, and they hang together:

    • The escaping tendency of both components increases, so the vapour pressure of the solution is greater than Raoult's law predicts. The solution therefore shows a positive deviation.
    • Breaking the stronger bonds costs more energy than forming the weaker ones releases, so mixing is endothermic and ${\Delta }_{\text{mix}}H>0$.
    • The looser packing means the total volume increases slightly, so ${\Delta }_{\text{mix}}V>0$.

    This is the standard illustration of a non-ideal solution with positive deviation, and the same reasoning explains why such mixtures can form minimum-boiling azeotropes.

    ✦ The new alcohol-water hydrogen bonds are weaker than those they replace, so vapour pressure rises, giving positive deviation from Raoult's law with positive enthalpy and volume of mixing

11. 1.112 marksNCERT Exercises, Chapter 1

    Why do gases always tend to be less soluble in liquids as the temperature is raised?

    Hint. Dissolution of a gas is exothermic, so apply Le Chatelier's principle to the equilibrium.

    Dissolving a gas in a liquid is an exothermic process, since the gas molecules are being brought closer together and attractive forces are established with the solvent:

    $\text{gas}+\text{solvent}\rightleftharpoons \text{solution}+\text{heat}$

    By Le Chatelier's principle, raising the temperature adds heat to a system that already releases heat in the forward direction, so the equilibrium shifts backwards. Dissolved gas is therefore expelled and the solubility falls.

    There is also a kinetic way of seeing the same thing: heating increases the kinetic energy of the dissolved molecules, so more of them acquire enough energy to escape the intermolecular attractions holding them in solution.

    This is why boiling water drives out dissolved air, and why warm water holds less dissolved oxygen, a fact of real importance for aquatic life in heated rivers.

    ✦ Because dissolution of a gas is exothermic, so by Le Chatelier's principle raising the temperature shifts the equilibrium backwards and expels dissolved gas

12. 1.123 marksNCERT Exercises, Chapter 1

    State Henry's law and mention some important applications.

    Hint. State the law in terms of partial pressure and mole fraction, then look for situations where dissolved gas pressure matters.

    Henry's law states that at constant temperature, the partial pressure of a gas in the vapour phase is directly proportional to its mole fraction in the solution:

    $p=K_{H}x$

    where $K_H$ is the Henry's law constant, characteristic of the gas and the solvent. A higher $K_H$ means a lower solubility, and $K_H$ increases with temperature, which is another way of saying gases become less soluble on heating.

    Applications:

    • Carbonated drinks. Soda bottles are sealed under high CO${}_2$ pressure to keep the gas dissolved. Opening the bottle drops the pressure and the gas fizzes out.
    • Deep-sea diving. At depth the high pressure dissolves more nitrogen in the blood. Surfacing too quickly releases it as bubbles, causing the painful and dangerous condition known as bends. Divers therefore use air diluted with helium, which is far less soluble.
    • High altitude. The low partial pressure of oxygen means less dissolves in the blood, causing the weakness and confused thinking of anoxia felt by climbers.
    • Medical oxygen. Patients with breathing difficulty are given oxygen-enriched air, since a higher partial pressure drives more oxygen into the blood.

    ✦ Henry's law states that p = K_H x at constant temperature; it explains carbonated drinks, the bends in deep-sea divers, anoxia at high altitude and the use of oxygen-rich air in medicine

13. 1.132 marksNCERT Exercises, Chapter 1

    The partial pressure of ethane over a solution containing $6.56\times 10^{-3}$ g of ethane is 1 bar. If the solution contains $5.00\times 10^{-2}$ g of ethane, then what shall be the partial pressure of the gas?

    Hint. In such a dilute solution the mole fraction is proportional to the mass dissolved, so the pressure scales directly with it.

    Henry's law gives $p=K_{H}x$. In a very dilute solution the mole fraction of the dissolved gas is proportional to the number of moles, which is in turn proportional to the mass, since the molar mass is fixed. So the partial pressure scales directly with the mass dissolved:

    $\frac{p_2}{p_1}=\frac{m_2}{m_1}$

    Substituting:

    $p_2=1\times \frac{5.00\times 10^{-2}}{6.56\times 10^{-3}}=7.62\text{bar}$

    There is no need to find $K_H$ itself or to know the molar mass of ethane, because both cancel in the ratio.

    ✦ 7.62 bar

14. 1.144 marksNCERT Exercises, Chapter 1

    What is meant by positive and negative deviations from Raoult's law and how is the sign of ${\Delta }_{\text{mix}}H$ related to positive and negative deviations from Raoult's law?

    Hint. Compare the strength of the A-B interaction in the mixture with the A-A and B-B interactions in the pure liquids.

    A solution deviates from Raoult's law when the interaction between unlike molecules differs in strength from that between like molecules.

    Positive deviation. The A-B attraction is weaker than the A-A and B-B attractions. Molecules escape more readily, so the observed vapour pressure is higher than Raoult's law predicts. Breaking the stronger original bonds costs more than the weaker new ones release, so mixing is endothermic:

    ${\Delta }_{\text{mix}}H>0,{\Delta }_{\text{mix}}V>0$

    Examples are ethanol with water, and acetone with carbon disulphide.

    Negative deviation. The A-B attraction is stronger than the A-A and B-B attractions. Molecules are held more tightly, so the vapour pressure is lower than predicted. Forming the stronger bonds releases more energy than breaking the weaker ones absorbs, so mixing is exothermic:

    ${\Delta }_{\text{mix}}H<0,{\Delta }_{\text{mix}}V<0$

    Examples are chloroform with acetone, where hydrogen bonding forms between the two, and nitric acid with water.

    For an ideal solution the two interactions are equal in strength, so ${\Delta }_{\text{mix}}H=0$ and ${\Delta }_{\text{mix}}V=0$. The sign of the enthalpy of mixing therefore tells you the direction of the deviation immediately.

    ✦ Positive deviation means weaker A-B interactions, higher vapour pressure and endothermic mixing with positive enthalpy of mixing; negative deviation means stronger A-B interactions, lower vapour pressure and exothermic mixing with negative enthalpy of mixing

15. 1.153 marksNCERT Exercises, Chapter 1

    An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

    Hint. At the normal boiling point the vapour pressure of the pure solvent is 1.013 bar, which is the value you need for the relative lowering.

    At the normal boiling point the vapour pressure of pure water is by definition 1 atm, that is $1.013$ bar. That is the datum the question expects you to supply.

    The relative lowering of vapour pressure equals the mole fraction of the solute:

    $\frac{p^{\circ }-p}{p^{\circ }}=\frac{1.013-1.004}{1.013}=\frac{0.009}{1.013}=8.885\times 10^{-3}$

    Take 100 g of solution, containing 2 g of solute and 98 g of water, so $n_{{\text{H}}_2\text{O}}=98/18=5.444$ mol. Since the solution is dilute, the solute moles may be neglected in the denominator of the mole fraction:

    $x_{\text{solute}}\approx \frac{n_{\text{solute}}}{n_{{\text{H}}_2\text{O}}}=\frac{2/M}{5.444}=8.885\times 10^{-3}$

    $\frac{2}{M}=0.04837⟹M=41.35{\text{g mol}}^{-1}$

    ✦ Molar mass = 41.35 g/mol

16. 1.163 marksNCERT Exercises, Chapter 1

    Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

    Hint. Convert both masses to moles, find the mole fractions, then apply Raoult's law for an ideal solution.

    With $M({\text{C}}_7{\text{H}}_{16})=100$ and $M({\text{C}}_8{\text{H}}_{18})=114$ g mol${}^{-1}$:

    $n_{\text{hep}}=\frac{26.0}{100}=0.260\text{mol},n_{\text{oct}}=\frac{35}{114}=0.307\text{mol}$

    The total is $0.567$ mol, so:

    $x_{\text{hep}}=\frac{0.260}{0.567}=0.4586,x_{\text{oct}}=0.5414$

    Since the solution is ideal, Raoult's law applies to both components:

    $p_{\text{total}}=p_{\text{hep}}^{\circ }{x}_{\text{hep}}+p_{\text{oct}}^{\circ }{x}_{\text{oct}}$

    $p_{\text{total}}=105.2\times 0.4586+46.8\times 0.5414=48.24+25.34=73.6\text{kPa}$

    ✦ 73.6 kPa

17. 1.172 marksNCERT Exercises, Chapter 1

    The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

    Hint. A 1 molal solution means 1 mole of solute in 1000 g of water; convert that into a mole fraction.

    A 1 molal solution contains 1 mol of solute in 1000 g of water, and:

    $n_{{\text{H}}_2\text{O}}=\frac{1000}{18}=55.55\text{mol}$

    The mole fraction of the solute is therefore:

    $x_{\text{solute}}=\frac{1}{1+55.55}=\frac{1}{56.55}=0.01768$

    Applying Raoult's law for a non-volatile solute:

    $p=p^{\circ }(1-x_{\text{solute}})=12.3\times (1-0.01768)=12.3\times 0.98232$

    $p=12.08\text{kPa}$

    The solute moles must be kept in the denominator here, since 1 mol against 55.55 mol is not entirely negligible.

    ✦ 12.08 kPa

18. 1.183 marksNCERT Exercises, Chapter 1

    Calculate the mass of a non-volatile solute (molar mass 40 g mol${}^{-1}$) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

    Hint. Reducing the vapour pressure to 80% means the mole fraction of the solvent is 0.80, by Raoult's law.

    By Raoult's law the vapour pressure of the solution is $p=p^{\circ }{x}_{\text{solvent}}$. Reducing it to 80% of the pure value means:

    $\frac{p}{p^{\circ }}=x_{\text{octane}}=0.80$

    With $M({\text{C}}_8{\text{H}}_{18})=114$ g mol${}^{-1}$, the 114 g of octane is exactly 1 mol. Letting $n$ be the moles of solute:

    $\frac{1}{1+n}=0.80⟹1=0.8+0.8n⟹n=0.25\text{mol}$

    The mass required is therefore:

    $0.25\times 40=10\text{g}$

    The numbers are chosen so that the octane works out to exactly one mole, which keeps the algebra clean.

    ✦ 10 g of solute

19. 1.195 marksNCERT Exercises, Chapter 1

    A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.

    Hint. Write Raoult's law for both situations, then divide one equation by the other to eliminate the unknown vapour pressure of pure water.

    Let $M$ be the molar mass of the solute and $p^{\circ }$ the vapour pressure of pure water. Write $a=30/M$ for the moles of solute, which is the same in both cases.

    Before adding water: $n_{{\text{H}}_2\text{O}}=90/18=5$ mol, so Raoult's law gives:

    $2.8=p^{\circ }\cdot \frac{5}{5+a}$

    After adding 18 g of water: $n_{{\text{H}}_2\text{O}}=108/18=6$ mol, so:

    $2.9=p^{\circ }\cdot \frac{6}{6+a}$

    Dividing the first by the second eliminates $p^{\circ }$:

    $\frac{2.8}{2.9}=\frac{5(6+a)}{6(5+a)}=\frac{30+5a}{30+6a}$

    $2.8(30+6a)=2.9(30+5a)$

    $84+16.8a=87+14.5a⟹2.3a=3⟹a=1.304$

    (i) Therefore:

    $M=\frac{30}{1.304}=23{\text{g mol}}^{-1}$

    (ii) Substituting back into the first equation:

    $p^{\circ }=2.8\times \frac{5+1.304}{5}=2.8\times 1.2609=3.53\text{kPa}$

    ✦ (i) Molar mass = 23 g/mol (ii) Vapour pressure of pure water = 3.53 kPa

20. 1.204 marksNCERT Exercises, Chapter 1

    A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

    Hint. Use the cane sugar data to find Kf for water first, then apply it to the glucose solution.

    The cane sugar data is there to supply $K_f$, which is then reused for glucose.

    Step 1: find $K_f$. For cane sugar, $\Delta T_f=273.15-271=2.15$ K. A 5% solution has 5 g of sucrose ($M=342$) in 95 g of water:

    $m=\frac{5/342}{0.095}=\frac{0.01462}{0.095}=0.1539{\text{mol kg}}^{-1}$

    $K_f=\frac{\Delta T_f}{m}=\frac{2.15}{0.1539}=13.97{\text{K kg mol}}^{-1}$

    Step 2: apply it to glucose. A 5% solution has 5 g of glucose ($M=180$) in 95 g of water:

    $m=\frac{5/180}{0.095}=\frac{0.02778}{0.095}=0.2924{\text{mol kg}}^{-1}$

    $\Delta T_f=13.97\times 0.2924=4.09\text{K}$

    $T_f=273.15-4.09=269.06\text{K}$

    The glucose solution freezes lower because glucose has the smaller molar mass, so the same 5 g by mass supplies almost twice as many particles.

    ✦ Freezing point = 269.06 K

21. 1.215 marksNCERT Exercises, Chapter 1

    Two elements A and B form compounds having formula AB${}_2$ and AB${}_4$. When dissolved in 20 g of benzene (C${}_6$H${}_6$), 1 g of AB${}_2$ lowers the freezing point by 2.3 K whereas 1.0 g of AB${}_4$ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol${}^{-1}$. Calculate atomic masses of A and B.

    Hint. Find the molar mass of each compound from its freezing point depression, then solve the two resulting equations together.

    Find each molar mass from $\Delta T_f=K_{f}m$, where $m=\frac{w/M}{W}$ with $w$ the solute mass and $W$ the solvent mass in kg. Rearranging:

    $M=\frac{K_{f}w}{\Delta T_f\times W}$

    For AB${}_2$:

    $M=\frac{5.1\times 1}{2.3\times 0.020}=\frac{5.1}{0.046}=110.87{\text{g mol}}^{-1}$

    For AB${}_4$:

    $M=\frac{5.1\times 1.0}{1.3\times 0.020}=\frac{5.1}{0.026}=196.15{\text{g mol}}^{-1}$

    Now write both in terms of the atomic masses:

    $A+2B=110.87\text{and}A+4B=196.15$

    Subtracting the first from the second:

    $2B=85.28⟹B=42.64$

    $A=110.87-2(42.64)=25.59$

    So the atomic mass of A is about 25.6 and that of B about 42.6.

    ✦ Atomic mass of A = 25.59, atomic mass of B = 42.64

22. 1.223 marksNCERT Exercises, Chapter 1

    At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

    Hint. At constant temperature osmotic pressure is directly proportional to concentration, so no value of R is needed.

    The van't Hoff equation is $\Pi =CRT$, so at a fixed temperature the osmotic pressure is directly proportional to the concentration. That means the ratio can be used and $R$ never appears.

    The original concentration, with $M(\text{glucose})=180$ g mol${}^{-1}$, is:

    $C_1=\frac{36/180}{1}=0.2{\text{mol L}}^{-1}$

    Applying the proportionality:

    $\frac{C_2}{C_1}=\frac{{\Pi }_2}{{\Pi }_1}⟹C_2=0.2\times \frac{1.52}{4.98}$

    $C_2=0.061{\text{mol L}}^{-1}$

    Working through $R$ and $T$ separately gives the same answer but takes far longer and invites arithmetic slips.

    ✦ 0.061 mol/L

23. 1.234 marksNCERT Exercises, Chapter 1

    Suggest the most important type of intermolecular attractive interaction in the following pairs: (i) n-hexane and n-octane (ii) I${}_2$ and CCl${}_4$ (iii) NaClO${}_4$ and water (iv) methanol and acetone (v) acetonitrile (CH${}_3$CN) and acetone (C${}_3$H${}_6$O).

    Hint. For each pair decide whether either species is ionic, and whether either can donate a hydrogen bond.

    (i) n-hexane and n-octane: both are non-polar hydrocarbons, so the only interaction available is London dispersion forces, which are significant here because both molecules are large and easily polarised.

    (ii) I${}_2$ and CCl${}_4$: iodine is non-polar and carbon tetrachloride is symmetrical and therefore non-polar overall, so again London dispersion forces.

    (iii) NaClO${}_4$ and water: sodium perchlorate is ionic and water is polar, so the interaction is ion-dipole. This is the strongest of the five and explains why the salt dissolves readily.

    (iv) Methanol and acetone: methanol has an O-H group and can donate a hydrogen bond, while acetone's carbonyl oxygen accepts one, so hydrogen bonding dominates, alongside weaker dipole-dipole forces.

    (v) Acetonitrile and acetone: both are polar but neither has a hydrogen attached to N, O or F, so no hydrogen bond can form. The interaction is dipole-dipole.

    The general rule of thumb: check for ions first, then for an O-H or N-H able to hydrogen bond, then fall back on dipole-dipole and finally dispersion.

    ✦ (i) London dispersion (ii) London dispersion (iii) ion-dipole (iv) hydrogen bonding (v) dipole-dipole

24. 1.243 marksNCERT Exercises, Chapter 1

    Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain: Cyclohexane, KCl, CH${}_3$OH, CH${}_3$CN.

    Hint. n-octane is non-polar, so apply the principle that like dissolves like.

    n-Octane is a non-polar hydrocarbon, so by the principle that like dissolves like, the most non-polar solute will be the most soluble.

    Ranking the four by polarity:

    • KCl is ionic. Dissolving it would require breaking strong ionic bonds with nothing but weak dispersion forces on offer in return, so it is essentially insoluble.
    • CH${}_3$OH is polar and strongly hydrogen bonded to itself. Those hydrogen bonds cannot be compensated in a hydrocarbon, so its solubility is low, though better than an ionic solid.
    • CH${}_3$CN is polar but cannot hydrogen bond, so less energy is lost on mixing and it dissolves better than methanol.
    • Cyclohexane is non-polar, like the solvent itself, so it is completely miscible.

    The increasing order of solubility is therefore:

    $\text{KCl}<{\text{CH}}_3\text{OH}<{\text{CH}}_3\text{CN}<\text{cyclohexane}$

    ✦ KCl < CH3OH < CH3CN < cyclohexane

25. 1.253 marksNCERT Exercises, Chapter 1

    Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water: (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol.

    Hint. Judge each by whether it can hydrogen bond with water and by how large its hydrocarbon portion is.

    Water is polar and strongly hydrogen bonded, so solubility depends on whether the solute can hydrogen bond with it, and on how much non-polar bulk it carries.

    Highly soluble:

    • Formic acid (HCOOH) — small and able to hydrogen bond through both the O-H and the C=O.
    • Ethylene glycol — two O-H groups on a two-carbon chain, so hydrogen bonding dominates completely.

    Partially soluble:

    • Phenol — the O-H hydrogen bonds with water, but the benzene ring is a substantial non-polar portion that resists dissolution.
    • Pentanol — the O-H helps, but a five-carbon chain is long enough to limit solubility considerably.

    Insoluble:

    • Toluene — a non-polar hydrocarbon with no capacity for hydrogen bonding.
    • Chloroform — although it has a small dipole, it cannot hydrogen bond effectively with water and is essentially immiscible.

    The pattern across the alcohols is worth remembering: solubility falls steadily as the hydrocarbon chain lengthens, because the non-polar part comes to outweigh the hydrogen-bonding group.

    ✦ Highly soluble: formic acid, ethylene glycol. Partially soluble: phenol, pentanol. Insoluble: toluene, chloroform

26. 1.263 marksNCERT Exercises, Chapter 1

    If the density of some lake water is 1.25 g mL${}^{-1}$ and contains 92 g of Na${}^{+}$ ions per kg of water, calculate the molarity of Na${}^{+}$ ions in the lake.

    Hint. The 92 g is per kg of water, not per kg of solution, so build the total mass before using the density.

    With the atomic mass of sodium as 23:

    $n_{{\text{Na}}^{+}}=\frac{92}{23}=4\text{mol}$

    The 92 g is stated per kilogram of water, so the total mass of solution is:

    $1000+92=1092\text{g}$

    The density then converts that into a volume:

    $V=\frac{1092}{1.25}=873.6\text{mL}=0.8736\text{L}$

    Therefore:

    $\text{Molarity}=\frac{4}{0.8736}=4.58{\text{mol L}}^{-1}$

    Using 1000 g as the mass of solution instead of 1092 g would give 5.0 M, which is the error the wording is set up to catch.

    ✦ 4.58 M

27. 1.272 marksNCERT Exercises, Chapter 1

    If the solubility product of CuS is $6\times 10^{-16}$, calculate the maximum molarity of CuS in aqueous solution.

    Hint. Write the dissociation equilibrium and express the solubility product in terms of the solubility s.

    Copper sulphide dissociates as:

    $\text{CuS}\rightleftharpoons {\text{Cu}}^{2+}+{\text{S}}^{2-}$

    If the solubility is $s$ mol L${}^{-1}$, then at saturation both ion concentrations equal $s$, so:

    $K_{sp}=[{\text{Cu}}^{2+}][{\text{S}}^{2-}]=s\times s=s^2$

    Therefore:

    $s=\sqrt{K_{sp}}=\sqrt{6\times 10^{-16}}=2.45\times 10^{-8}{\text{mol L}}^{-1}$

    The maximum molarity of dissolved CuS is therefore about $2.45\times 10^{-8}$ M, which is extremely small. This is why copper sulphide is precipitated so completely in qualitative analysis.

    ✦ 2.45 x 10^-8 mol/L

28. 1.282 marksNCERT Exercises, Chapter 1

    Calculate the mass percentage of aspirin (C${}_9$H${}_8$O${}_4$) in acetonitrile (CH${}_3$CN) when 6.5 g of C${}_9$H${}_8$O${}_4$ is dissolved in 450 g of CH${}_3$CN.

    Hint. Divide by the total mass of solution, which includes the aspirin itself.

    The total mass of the solution is the sum of solute and solvent:

    $6.5+450=456.5\text{g}$

    The mass percentage of aspirin is therefore:

    $\%=\frac{6.5}{456.5}\times 100=1.424\%$

    Dividing by 450 g instead of 456.5 g would give 1.444%, so it matters that the denominator is the mass of the whole solution and not of the solvent alone.

    ✦ 1.424%

29. 1.294 marksNCERT Exercises, Chapter 1

    Nalorphene (C${}_{19}$H${}_{21}$NO${}_3$), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of $1.5\times 10^{-3}$ m aqueous solution required for the above dose.

    Hint. Convert the dose to moles, then use the molality to find the mass of water that must accompany it.

    First find the molar mass of nalorphene, C${}_{19}$H${}_{21}$NO${}_3$:

    $19(12)+21(1)+14+3(16)=228+21+14+48=311{\text{g mol}}^{-1}$

    The dose of 1.5 mg is:

    $n=\frac{1.5\times 10^{-3}}{311}=4.823\times 10^{-6}\text{mol}$

    Molality is moles of solute per kilogram of solvent, so:

    $\text{mass of water}=\frac{4.823\times 10^{-6}}{1.5\times 10^{-3}}=3.215\times 10^{-3}\text{kg}=3.215\text{g}$

    The mass of solution is the water plus the nalorphene itself:

    $3.215+0.0015=3.217\text{g}\approx 3.22\text{g}$

    The solute contributes almost nothing to the total, because the solution is so dilute.

    ✦ About 3.22 g of solution

30. 1.302 marksNCERT Exercises, Chapter 1

    Calculate the amount of benzoic acid (C${}_6$H${}_5$COOH) required for preparing 250 mL of 0.15 M solution in methanol.

    Hint. Molarity times volume in litres gives the moles directly.

    The number of moles follows from the definition of molarity:

    $n=M\times V=0.15\times 0.250=0.0375\text{mol}$

    The molar mass of benzoic acid, C${}_6$H${}_5$COOH or C${}_7$H${}_6$O${}_2$, is:

    $7(12)+6(1)+2(16)=84+6+32=122{\text{g mol}}^{-1}$

    Therefore:

    $\text{mass}=0.0375\times 122=4.575\text{g}$

    Note that the volume must be converted to litres before multiplying, since molarity is defined per litre.

    ✦ 4.575 g of benzoic acid

31. 1.313 marksNCERT Exercises, Chapter 1

    The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

    Hint. Depression depends on the number of particles, so consider how far each acid dissociates and what controls that.

    Freezing point depression is a colligative property, so it depends on the number of particles in solution. For these three acids the number of particles depends on how completely each dissociates.

    All three are carboxylic acids, and dissociation is promoted by the electron-withdrawing inductive effect of the substituents, which stabilises the resulting carboxylate anion.

    • Acetic acid (CH${}_3$COOH) has a methyl group, which is electron releasing. It destabilises the anion, so dissociation is least and the fewest particles are produced.
    • Trichloroacetic acid (CCl${}_3$COOH) has three chlorine atoms, which withdraw electron density and stabilise the anion, so it dissociates far more.
    • Trifluoroacetic acid (CF${}_3$COOH) has three fluorine atoms. Since fluorine is more electronegative than chlorine, the withdrawing effect is stronger still and dissociation is greatest.

    More dissociation means more particles, a larger van't Hoff factor $i$, and therefore a greater depression:

    ${\text{CH}}_3\text{COOH}<{\text{CCl}}_3\text{COOH}<{\text{CF}}_3\text{COOH}$

    ✦ Because the electron-withdrawing power of the substituents increases in that order, so dissociation and hence the number of particles increases, giving a larger van't Hoff factor and a greater depression

32. 1.324 marksNCERT Exercises, Chapter 1

    Calculate the depression in the freezing point of water when 10 g of CH${}_3$CH${}_2$CHClCOOH is added to 250 g of water. $K_a=1.4\times 10^{-3}$, $K_f=1.86$ K kg mol${}^{-1}$.

    Hint. The acid partially dissociates, so find the degree of dissociation from Ka first and use it to build the van't Hoff factor.

    The molar mass of CH${}_3$CH${}_2$CHClCOOH, which is C${}_4$H${}_7$ClO${}_2$, is:

    $4(12)+7(1)+35.5+2(16)=48+7+35.5+32=122.5{\text{g mol}}^{-1}$

    $n=\frac{10}{122.5}=0.0816\text{mol},m=\frac{0.0816}{0.250}=0.3265{\text{mol kg}}^{-1}$

    Degree of dissociation. For the weak acid HA $\rightleftharpoons$ H${}^{+}$ + A${}^{-}$, with $\alpha$ small:

    $K_a=\frac{c{\alpha }^2}{1-\alpha }\approx c{\alpha }^2⟹\alpha =\sqrt{\frac{K_a}{c}}=\sqrt{\frac{1.4\times 10^{-3}}{0.3265}}=0.0655$

    Van't Hoff factor. One molecule gives two particles when it dissociates, so:

    $i=1+\alpha =1.0655$

    Depression:

    $\Delta T_f=i{K}_{f}m=1.0655\times 1.86\times 0.3265=0.65\text{K}$

    Ignoring the dissociation entirely would give 0.61 K, so the acid's partial ionisation makes a real difference.

    ✦ Depression = 0.65 K

33. 1.335 marksNCERT Exercises, Chapter 1

    19.5 g of CH${}_2$FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00${}^{\circ }$C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

    Hint. Compare the observed depression with the one calculated assuming no dissociation to get i, then work back to Ka.

    The molar mass of CH${}_2$FCOOH, that is C${}_2$H${}_3$FO${}_2$, is:

    $2(12)+3(1)+19+2(16)=24+3+19+32=78{\text{g mol}}^{-1}$

    $n=\frac{19.5}{78}=0.25\text{mol},m=\frac{0.25}{0.500}=0.5{\text{mol kg}}^{-1}$

    Calculated depression, assuming no dissociation:

    $\Delta T_f(\text{calc})=K_{f}m=1.86\times 0.5=0.93\text{K}$

    Van't Hoff factor is the ratio of observed to calculated:

    $i=\frac{\Delta T_f(\text{obs})}{\Delta T_f(\text{calc})}=\frac{1.00}{0.93}=1.0753$

    Since $i>1$, the acid dissociates, and $i=1+\alpha$ gives:

    $\alpha =0.0753$

    Dissociation constant:

    $K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.5\times (0.0753{)}^2}{1-0.0753}=\frac{0.5\times 5.670\times 10^{-3}}{0.9247}$

    $K_a=3.07\times 10^{-3}$

    ✦ van't Hoff factor i = 1.0753; Ka = 3.07 x 10^-3

34. 1.343 marksNCERT Exercises, Chapter 1

    Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

    Hint. Glucose is non-volatile, so apply Raoult's law using the mole fraction of the water.

    Glucose is non-volatile, so only the water contributes to the vapour pressure and Raoult's law gives $p=p^{\circ }{x}_{\text{water}}$.

    With $M(\text{glucose})=180$ g mol${}^{-1}$:

    $n_{\text{glucose}}=\frac{25}{180}=0.1389\text{mol},n_{{\text{H}}_2\text{O}}=\frac{450}{18}=25\text{mol}$

    $x_{\text{glucose}}=\frac{0.1389}{0.1389+25}=\frac{0.1389}{25.1389}=5.525\times 10^{-3}$

    Therefore:

    $p=p^{\circ }(1-x_{\text{glucose}})=17.535\times (1-5.525\times 10^{-3})$

    $p=17.535\times 0.994475=17.44\text{mm Hg}$

    The lowering is small, only about 0.1 mm Hg, because the solution is dilute.

    ✦ 17.44 mm Hg

35. 1.352 marksNCERT Exercises, Chapter 1

    Henry's law constant for the molality of methane in benzene at 298 K is $4.27\times 10^5$ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

    Hint. The constant is given for the molality form of Henry's law, so the answer comes out directly as a molality.

    The constant is quoted for the molality form of Henry's law, so the relation is:

    $p=K_{H}m$

    Rearranging for the solubility:

    $m=\frac{p}{K_H}=\frac{760}{4.27\times 10^5}$

    $m=1.78\times 10^{-3}{\text{mol kg}}^{-1}$

    Because the constant already refers to molality, no conversion into mole fraction is needed here. Reading the units of $K_H$ carefully before substituting is what keeps this question to a single line.

    ✦ 1.78 x 10^-3 mol/kg

36. 1.364 marksNCERT Exercises, Chapter 1

    100 g of liquid A (molar mass 140 g mol${}^{-1}$) was dissolved in 1000 g of liquid B (molar mass 180 g mol${}^{-1}$). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

    Hint. Find the mole fractions first, then use Raoult's law for the total pressure with the unknown as the vapour pressure of pure A.

    Find the moles of each component:

    $n_A=\frac{100}{140}=0.7143\text{mol},n_B=\frac{1000}{180}=5.5556\text{mol}$

    The total is $6.2699$ mol, so:

    $x_A=\frac{0.7143}{6.2699}=0.1139,x_B=0.8861$

    Applying Raoult's law to the total pressure:

    $p_{\text{total}}=p_A^{\circ }{x}_A+p_B^{\circ }{x}_B$

    $475=p_A^{\circ }(0.1139)+500(0.8861)=0.1139p_A^{\circ }+443.05$

    $0.1139p_A^{\circ }=31.95⟹p_A^{\circ }=280.6\text{torr}$

    The partial pressure of A in the solution is then:

    $p_A=p_A^{\circ }{x}_A=280.6\times 0.1139=31.96\text{torr}$

    Note that A is the more volatile component here, since its pure vapour pressure is lower than B's only in appearance; what matters for the total is the product with the mole fraction, and A's small mole fraction is why it contributes so little.

    ✦ Vapour pressure of pure A = 280.6 torr; its partial pressure in the solution = 31.96 torr

37. 1.374 marksNCERT Exercises, Chapter 1

    Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot $p_{\text{total}}$, $p_{\text{chloroform}}$ and $p_{\text{acetone}}$ as a function of $x_{\text{acetone}}$. The experimental data observed for different compositions of mixture is given in the textbook. Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

    Hint. Compare the experimental total pressures with the straight line predicted by Raoult's law.

    The ideal plot. For an ideal solution Raoult's law gives three straight lines against $x_{\text{acetone}}$:

    $p_{\text{acetone}}=741.8x_{\text{acetone}}$

    $p_{\text{chloroform}}=632.8(1-x_{\text{acetone}})$

    $p_{\text{total}}=632.8+109x_{\text{acetone}}$

    So the ideal total pressure rises linearly from 632.8 mm Hg at $x=0$ to 741.8 mm Hg at $x=1$.

    The experimental data. Adding the two measured partial pressures at each composition gives a total that lies below this straight line throughout. For example at $x_{\text{acetone}}=0.234$ the measured total is $110.1+469.4=579.5$ mm Hg, whereas the ideal value would be $632.8+109(0.234)=658.3$ mm Hg.

    Conclusion: negative deviation. The experimental curve lies below the Raoult's law line at every composition, so the mixture shows negative deviation.

    Why. Acetone's carbonyl oxygen forms a hydrogen bond with the hydrogen of chloroform. This new interaction between unlike molecules is stronger than the acetone-acetone and chloroform-chloroform interactions it replaces, so molecules escape less readily and the vapour pressure falls below the ideal value. Consistently, mixing is exothermic with ${\Delta }_{\text{mix}}H<0$ and the volume contracts.

    ✦ The experimental total pressure lies below the Raoult's law straight line at every composition, so the mixture shows negative deviation, caused by hydrogen bonding between acetone and chloroform

38. 1.383 marksNCERT Exercises, Chapter 1

    Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

    Hint. Find the liquid mole fractions, then the two partial pressures, and finally apply Dalton's law for the vapour.

    With $M({\text{C}}_6{\text{H}}_6)=78$ and $M({\text{C}}_7{\text{H}}_8)=92$ g mol${}^{-1}$:

    $n_{\text{benzene}}=\frac{80}{78}=1.0256\text{mol},n_{\text{toluene}}=\frac{100}{92}=1.0870\text{mol}$

    The total is $2.1126$ mol, so in the liquid:

    $x_{\text{benzene}}=0.4855,x_{\text{toluene}}=0.5145$

    The partial pressures follow from Raoult's law:

    $p_{\text{benzene}}=50.71\times 0.4855=24.62\text{mm Hg}$

    $p_{\text{toluene}}=32.06\times 0.5145=16.50\text{mm Hg}$

    $p_{\text{total}}=41.12\text{mm Hg}$

    By Dalton's law the mole fraction in the vapour is the partial pressure over the total:

    $y_{\text{benzene}}=\frac{24.62}{41.12}=0.60$

    The vapour is richer in benzene (0.60) than the liquid was (0.4855), because benzene is the more volatile component.

    ✦ Mole fraction of benzene in the vapour phase = 0.60

39. 1.394 marksNCERT Exercises, Chapter 1

    The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry's law constants for oxygen and nitrogen at 298 K are $3.30\times 10^7$ mm and $6.51\times 10^7$ mm respectively, calculate the composition of these gases in water.

    Hint. Percentage by volume equals mole percentage for gases, so the partial pressures follow directly from the total pressure.

    For gases, percentage by volume equals mole percentage, so the partial pressures are simply fractions of the total:

    $p_{{\text{O}}_2}=0.20\times 10=2\text{atm}=2\times 760=1520\text{mm Hg}$

    $p_{{\text{N}}_2}=0.79\times 10=7.9\text{atm}=7.9\times 760=6004\text{mm Hg}$

    Applying Henry's law $x=p/K_H$ to each:

    $x_{{\text{O}}_2}=\frac{1520}{3.30\times 10^7}=4.61\times 10^{-5}$

    $x_{{\text{N}}_2}=\frac{6004}{6.51\times 10^7}=9.22\times 10^{-5}$

    So dissolved nitrogen exceeds dissolved oxygen by about a factor of two, even though nitrogen is the less soluble gas, because its partial pressure in air is nearly four times greater.

    ✦ x(O2) = 4.61 x 10^-5 and x(N2) = 9.22 x 10^-5

40. 1.403 marksNCERT Exercises, Chapter 1

    Determine the amount of CaCl${}_2$ ($i=2.47$) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27${}^{\circ }$C.

    Hint. Calcium chloride dissociates, so include the van't Hoff factor in the osmotic pressure equation.

    Because calcium chloride dissociates into ions, the van't Hoff factor must be included:

    $\Pi =i\frac{n}{V}RT$

    Rearranging for the number of moles, with $T=273+27=300$ K:

    $n=\frac{\Pi V}{iRT}=\frac{0.75\times 2.5}{2.47\times 0.0821\times 300}$

    $n=\frac{1.875}{60.84}=0.0308\text{mol}$

    The molar mass of CaCl${}_2$ is $40+2(35.5)=111$ g mol${}^{-1}$, so:

    $\text{mass}=0.0308\times 111=3.42\text{g}$

    Omitting $i$ would give 8.45 g, which is more than twice too large, since the dissociation supplies nearly 2.5 particles for every formula unit dissolved.

    ✦ About 3.42 g of CaCl2

41. 1.413 marksNCERT Exercises, Chapter 1

    Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K${}_2$SO${}_4$ in 2 litre of water at 25${}^{\circ }$C, assuming that it is completely dissociated.

    Hint. Complete dissociation of K2SO4 gives three ions per formula unit, which fixes the van't Hoff factor.

    Potassium sulphate dissociates completely as:

    ${\text{K}}_2{\text{SO}}_4\to 2{\text{K}}^{+}+{\text{SO}}_4^{2-}$

    One formula unit gives three ions, so $i=3$.

    The molar mass is $2(39)+32+4(16)=78+32+64=174$ g mol${}^{-1}$, so:

    $n=\frac{25\times 10^{-3}}{174}=1.437\times 10^{-4}\text{mol}$

    Applying the van't Hoff equation with $T=273+25=298$ K:

    $\Pi =i\frac{n}{V}RT=3\times \frac{1.437\times 10^{-4}}{2}\times 0.0821\times 298$

    $\Pi =5.27\times 10^{-3}\text{atm}$

    The pressure is small because the solution is very dilute, but osmotic pressure is still the easiest colligative property to measure at such low concentrations.

    ✦ 5.27 x 10^-3 atm

Solutions written by the tuition.in editorial team and checked against lech101.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.