Intext Questions — Solutions

1. 1.12 marksNCERT Intext Questions, Chapter 1

   Calculate the mass percentage of benzene (C${}_6$H${}_6$) and carbon tetrachloride (CCl${}_4$) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

   Hint. Mass percentage is the mass of a component divided by the total mass of the solution, not by the mass of the other component.

   The total mass of the solution is the sum of both components:

   $22+122=144\text{g}$

   The mass percentage of benzene is therefore:

   $\%{\text{C}}_6{\text{H}}_6=\frac{22}{144}\times 100=15.28\%$

   and for carbon tetrachloride:

   $\%{\text{CCl}}_4=\frac{122}{144}\times 100=84.72\%$

   As a check the two percentages sum to 100, which they must, since every gram of solution belongs to one component or the other.

   ✦ Benzene 15.28%, carbon tetrachloride 84.72%

2. 1.22 marksNCERT Intext Questions, Chapter 1

   Calculate the mole fraction of benzene in a solution containing 30% by mass in carbon tetrachloride.

   Hint. Take a convenient 100 g of solution so the percentages become masses directly, then convert each to moles.

   Take 100 g of solution, so that the masses are 30 g of benzene and 70 g of carbon tetrachloride.

   Converting to moles, with $M({\text{C}}_6{\text{H}}_6)=78$ and $M({\text{CCl}}_4)=154$ g mol${}^{-1}$:

   $n_{{\text{C}}_6{\text{H}}_6}=\frac{30}{78}=0.3846\text{mol},n_{{\text{CCl}}_4}=\frac{70}{154}=0.4545\text{mol}$

   The total is $0.8391$ mol, so:

   $x_{{\text{C}}_6{\text{H}}_6}=\frac{0.3846}{0.8391}=0.459$

   $x_{{\text{CCl}}_4}=1-0.459=0.541$

   Note that the mole fraction of benzene (0.459) exceeds its mass fraction (0.30), because benzene has the smaller molar mass and so contributes more particles per gram.

   ✦ x(benzene) = 0.459, x(carbon tetrachloride) = 0.541

3. 1.33 marksNCERT Intext Questions, Chapter 1

   Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO${}_3$)${}_2\cdot$6H${}_2$O in 4.3 L of solution (b) 30 mL of 0.5 M H${}_2$SO${}_4$ diluted to 500 mL.

   Hint. For (a) remember to include the water of crystallisation in the molar mass; for (b) use the fact that dilution leaves the number of moles unchanged.

   (a) The molar mass must include the six waters of crystallisation:

   $M=58.93+2(62.01)+6(18.02)=291.03{\text{g mol}}^{-1}$

   $n=\frac{30}{291.03}=0.1031\text{mol}$

   $\text{Molarity}=\frac{0.1031}{4.3}=0.024\text{M}$

   Omitting the water of crystallisation would give a molar mass of 183 and a badly wrong answer, so it is worth writing the full formula out first.

   (b) Dilution changes the volume but not the amount of solute, so $M_1{V}_1=M_2{V}_2$:

   $0.5\times 30=M_2\times 500$

   $M_2=\frac{15}{500}=0.03\text{M}$

   ✦ (a) 0.024 M (b) 0.03 M

4. 1.43 marksNCERT Intext Questions, Chapter 1

   Calculate the mass of urea (NH${}_2$CONH${}_2$) required in making 2.5 kg of 0.25 molal aqueous solution.

   Hint. Molality is defined per kilogram of solvent, not per kilogram of solution, so the 2.5 kg has to be split between the two.

   The trap here is that molality is defined per kilogram of solvent, while the 2.5 kg given is the mass of the whole solution.

   Let the mass of water be $w$ kg. A $0.25$ molal solution then contains $0.25w$ mol of urea, and with $M(\text{urea})=60$ g mol${}^{-1}$ that is a mass of $15w$ g, or $0.015w$ kg.

   The total mass of solution is the sum of the two:

   $w+0.015w=2.5⟹1.015w=2.5$

   $w=2.4631\text{kg}$

   The mass of urea is therefore:

   $0.015\times 2.4631\times 1000=36.95\text{g}$

   Treating the 2.5 kg as solvent would give 37.5 g, which is close enough to look right and is exactly the error the question is designed to catch.

   ✦ 36.95 g of urea

5. 1.54 marksNCERT Intext Questions, Chapter 1

   Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL${}^{-1}$.

   Hint. Take 100 g of solution so the percentage converts directly into masses, then use the density only for the molarity.

   Take 100 g of solution, giving 20 g of KI and 80 g of water. With $M(\text{KI})=166$ g mol${}^{-1}$:

   $n_{\text{KI}}=\frac{20}{166}=0.1205\text{mol}$

   (a) Molality uses the mass of solvent alone, which is $0.080$ kg:

   $m=\frac{0.1205}{0.080}=1.51{\text{mol kg}}^{-1}$

   (b) Molarity needs the volume, which is where the density enters:

   $V=\frac{100}{1.202}=83.19\text{mL}=0.08319\text{L}$

   $M=\frac{0.1205}{0.08319}=1.45{\text{mol L}}^{-1}$

   (c) Mole fraction needs the moles of water as well:

   $n_{{\text{H}}_2\text{O}}=\frac{80}{18}=4.444\text{mol}$

   $x_{\text{KI}}=\frac{0.1205}{0.1205+4.444}=0.0264$

   Notice that molality and molarity differ here, because the solution is concentrated and its density is not 1 g mL${}^{-1}$.

   ✦ (a) 1.51 mol/kg (b) 1.45 mol/L (c) 0.0264

6. 1.63 marksNCERT Intext Questions, Chapter 1

   H${}_2$S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H${}_2$S in water at STP is 0.195 m, calculate Henry's law constant.

   Hint. Convert the molality into a mole fraction first, since Henry's law in this form is written in terms of mole fraction.

   Henry's law states that the partial pressure of a gas is proportional to its mole fraction in solution:

   $p=K_{H}x$

   So the molality must first be converted into a mole fraction. A $0.195$ molal solution contains $0.195$ mol of H${}_2$S in 1000 g of water, and:

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

   $x_{{\text{H}}_2\text{S}}=\frac{0.195}{0.195+55.55}=\frac{0.195}{55.75}=3.498\times 10^{-3}$

   At STP the pressure is 1 bar, so:

   $K_H=\frac{p}{x}=\frac{1}{3.498\times 10^{-3}}=285.9\text{bar}$

   Expressed in atmospheres this is $282$ atm, since 1 bar is 0.987 atm. Both figures appear in solution books, so state which unit you are working in.

   ✦ K_H = 285.9 bar, equivalently about 282 atm

7. 1.73 marksNCERT Intext Questions, Chapter 1

   Henry's law constant for CO${}_2$ in water is $1.67\times 10^8$ Pa at 298 K. Calculate the quantity of CO${}_2$ in 500 mL of soda water when packed under 2.5 atm CO${}_2$ pressure at 298 K.

   Hint. Find the mole fraction from Henry's law, then use the fact that the solution is very dilute to simplify the conversion to moles.

   Convert the pressure to pascals so that it matches the units of $K_H$:

   $p=2.5\times 101325=2.533\times 10^5\text{Pa}$

   From Henry's law:

   $x_{{\text{CO}}_2}=\frac{p}{K_H}=\frac{2.533\times 10^5}{1.67\times 10^8}=1.517\times 10^{-3}$

   Taking 500 mL of water as 500 g:

   $n_{{\text{H}}_2\text{O}}=\frac{500}{18}=27.78\text{mol}$

   Because the mole fraction is so small, the moles of CO${}_2$ are negligible in the denominator, so:

   $n_{{\text{CO}}_2}\approx x\times n_{{\text{H}}_2\text{O}}=1.517\times 10^{-3}\times 27.78=0.0421\text{mol}$

   $\text{mass}=0.0421\times 44=1.85\text{g}$

   ✦ About 1.85 g of CO2

8. 1.84 marksNCERT Intext Questions, Chapter 1

   The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

   Hint. Write the total pressure using Raoult's law with a single unknown, then use Dalton's law for the vapour composition.

   For an ideal binary solution, Raoult's law gives the total pressure as:

   $p_{\text{total}}=p_A^{\circ }{x}_A+p_B^{\circ }(1-x_A)$

   Substituting the given values:

   $600=450x_A+700(1-x_A)=700-250x_A$

   $250x_A=100⟹x_A=0.40,x_B=0.60$

   For the vapour composition, use Dalton's law, since the mole fraction of a component in the vapour is its partial pressure divided by the total:

   $p_A=450\times 0.40=180\text{mm Hg},p_B=700\times 0.60=420\text{mm Hg}$

   $y_A=\frac{180}{600}=0.30,y_B=\frac{420}{600}=0.70$

   The vapour is richer in B, the more volatile component, which is the general rule and the basis of fractional distillation.

   ✦ Liquid: x(A) = 0.40, x(B) = 0.60. Vapour: y(A) = 0.30, y(B) = 0.70

9. 1.93 marksNCERT Intext Questions, Chapter 1

   Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH${}_2$CONH${}_2$) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

   Hint. The relative lowering of vapour pressure equals the mole fraction of the solute, which is Raoult's law for a non-volatile solute.

   Urea is non-volatile, so Raoult's law in the form of relative lowering applies:

   $\frac{p^{\circ }-p}{p^{\circ }}=x_{\text{solute}}$

   First find the moles, with $M(\text{urea})=60$ g mol${}^{-1}$:

   $n_{\text{urea}}=\frac{50}{60}=0.8333\text{mol},n_{{\text{H}}_2\text{O}}=\frac{850}{18}=47.22\text{mol}$

   $x_{\text{urea}}=\frac{0.8333}{0.8333+47.22}=\frac{0.8333}{48.06}=0.01734$

   So the relative lowering is 0.0173, and the vapour pressure of the solution is:

   $p=p^{\circ }(1-x_{\text{urea}})=23.8\times 0.98266=23.4\text{mm Hg}$

   ✦ Vapour pressure = 23.4 mm Hg; relative lowering = 0.0173

10. 1.103 marksNCERT Intext Questions, Chapter 1

    Boiling point of water at 750 mm Hg is 99.63${}^{\circ }$C. How much sucrose is to be added to 500 g of water such that it boils at 100${}^{\circ }$C?

    Hint. The elevation is measured from the boiling point of pure water at the same pressure, which is 99.63 degrees here, not 100.

    The elevation in boiling point is measured from the boiling point of the pure solvent at that pressure, which the question gives as $99.63^{\circ }$C:

    $\Delta T_b=100-99.63=0.37\text{K}$

    Using $\Delta T_b=K_{b}m$ with $K_b=0.52$ K kg mol${}^{-1}$ for water:

    $m=\frac{0.37}{0.52}=0.7115{\text{mol kg}}^{-1}$

    In 500 g, that is 0.5 kg, of water:

    $n=0.7115\times 0.5=0.3558\text{mol}$

    With $M(\text{sucrose})=342$ g mol${}^{-1}$:

    $\text{mass}=0.3558\times 342=121.7\text{g}$

    Taking the reference as $100^{\circ }$C would give a zero elevation and no answer at all, so the 750 mm Hg detail is doing real work.

    ✦ About 121.7 g of sucrose

11. 1.113 marksNCERT Intext Questions, Chapter 1

    Calculate the mass of ascorbic acid (Vitamin C, C${}_6$H${}_8$O${}_6$) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5${}^{\circ }$C. $K_f=3.9$ K kg mol${}^{-1}$.

    Hint. Depression of freezing point and lowering of melting point are the same thing, so apply the standard relation directly.

    Lowering the melting point by $1.5^{\circ }$C is the same as a freezing point depression of $\Delta T_f=1.5$ K, since melting and freezing occur at the same temperature.

    From $\Delta T_f=K_{f}m$:

    $m=\frac{1.5}{3.9}=0.3846{\text{mol kg}}^{-1}$

    In 75 g, that is 0.075 kg, of acetic acid:

    $n=0.3846\times 0.075=0.02885\text{mol}$

    The molar mass of ascorbic acid, C${}_6$H${}_8$O${}_6$, is $6(12)+8(1)+6(16)=176$ g mol${}^{-1}$, so:

    $\text{mass}=0.02885\times 176=5.08\text{g}$

    Note that $K_f$ here belongs to acetic acid, not water, because acetic acid is the solvent in this problem.

    ✦ About 5.08 g of ascorbic acid

12. 1.123 marksNCERT Intext Questions, Chapter 1

    Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37${}^{\circ }$C.

    Hint. Use the van't Hoff equation, and convert the final answer from atmospheres into pascals.

    The osmotic pressure of a dilute solution is given by the van't Hoff equation:

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

    The number of moles of polymer is very small because the molar mass is enormous:

    $n=\frac{1.0}{185000}=5.405\times 10^{-6}\text{mol}$

    With $V=0.450$ L, $R=0.0821$ L atm K${}^{-1}$ mol${}^{-1}$ and $T=273+37=310$ K:

    $\Pi =\frac{5.405\times 10^{-6}\times 0.0821\times 310}{0.450}=3.06\times 10^{-4}\text{atm}$

    Converting to pascals, using 1 atm $=101325$ Pa:

    $\Pi =3.06\times 10^{-4}\times 101325=30.96\text{Pa}$

    This tiny but measurable pressure is why osmotic pressure is the colligative property of choice for finding the molar masses of polymers and proteins, since the other three would give changes far too small to detect.

    ✦ About 30.96 Pa

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.