Solutions: Numerical Practice

Comprehensive problem set covering concentration terms, Raoult's law, colligative properties, and van't Hoff factor.

TOPIC 1: Concentration of Solutions

Q1. Molarity & Molality CBSE

A solution of glucose in water is labelled as $10\%\text{ w/w}$. If the density of the solution is $1.20\text{ g mL}^{-1}$, calculate the molarity and molality of the solution.

Solution:

$10\%\text{ w/w}$ means $10\text{ g}$ glucose ($\text{C}_6\text{H}_{12}\text{O}_6$, $M=180\text{ g/mol}$) in $100\text{ g}$ solution.

Mass of solvent (water) $= 100\text{ g} - 10\text{ g} = 90\text{ g} = 0.09\text{ kg}$.

Moles of glucose ($n$) $= \frac{10}{180} = 0.0555\text{ mol}$.

Molality ($m$) $= \frac{n}{\text{Mass of solvent (kg)}} = \frac{0.0555}{0.09} = \mathbf{0.617\text{ m}}$.

Volume of solution $= \frac{\text{Mass}}{\text{Density}} = \frac{100\text{ g}}{1.20\text{ g mL}^{-1}} = 83.33\text{ mL} = 0.0833\text{ L}$.

Molarity ($M$) $= \frac{n}{\text{Volume (L)}} = \frac{0.0555}{0.0833} = \mathbf{0.666\text{ M}}$.

Q2. Mole Fraction CBSE

Calculate the mole fraction of ethylene glycol ($\text{C}_2\text{H}_6\text{O}_2$) in a solution containing $20\%$ of $\text{C}_2\text{H}_6\text{O}_2$ by mass.

Solution:

Assume $100\text{ g}$ of solution.

Mass of $\text{C}_2\text{H}_6\text{O}_2$ ($w_{\text{A}}$) $= 20\text{ g}$. Molar mass ($M_{\text{A}}$) $= 62\text{ g/mol}$.

Mass of water ($w_{\text{B}}$) $= 80\text{ g}$. Molar mass ($M_{\text{B}}$) $= 18\text{ g/mol}$.

Moles of A ($n_{\text{A}}$) $= \frac{20}{62} = 0.322\text{ mol}$.

Moles of B ($n_{\text{B}}$) $= \frac{80}{18} = 4.444\text{ mol}$.

Total moles $= 0.322 + 4.444 = 4.766\text{ mol}$.

Mole fraction of A ($x_{\text{A}}$) $= \frac{n_{\text{A}}}{\text{Total moles}} = \frac{0.322}{4.766} = \mathbf{0.068}$.

Q3. Mixing Solutions JEE Mains

What is the molarity of a solution resulting from mixing $2.5\text{ L}$ of $0.5\text{ M}$ urea solution and $500\text{ mL}$ of $2\text{ M}$ urea solution?

Solution:

Use the formula: $M_{\text{mix}}V_{\text{mix}} = M_1V_1 + M_2V_2$

$V_1 = 2.5\text{ L}$, $M_1 = 0.5\text{ M}$.

$V_2 = 500\text{ mL} = 0.5\text{ L}$, $M_2 = 2\text{ M}$.

Total Volume ($V_{\text{mix}}$) $= 2.5 + 0.5 = 3.0\text{ L}$.

$M_{\text{mix}} \times 3.0 = (0.5 \times 2.5) + (2 \times 0.5)$

$M_{\text{mix}} \times 3.0 = 1.25 + 1.0 = 2.25$

$M_{\text{mix}} = \frac{2.25}{3.0} = \mathbf{0.75\text{ M}}$.

Q4. Interconversion JEE Mains

The mole fraction of a solute in a $1.00\text{ molal}$ aqueous solution is?

Solution:

$1.00\text{ molal}$ aqueous solution means $1\text{ mole}$ of solute is dissolved in $1\text{ kg}$ ($1000\text{ g}$) of water.

Moles of solute ($n_2$) $= 1\text{ mol}$.

Mass of water $= 1000\text{ g}$. Moles of water ($n_1$) $= \frac{1000}{18} = 55.55\text{ mol}$.

Mole fraction of solute ($x_2$) $= \frac{n_2}{n_1 + n_2} = \frac{1}{55.55 + 1} = \frac{1}{56.55} = \mathbf{0.0177}$.

Q5. PPM Calculation CBSE

A sample of drinking water was found to be severely contaminated with chloroform ($\text{CHCl}_3$) supposed to be a carcinogen. The level of contamination was $15\text{ ppm}$ (by mass). Determine its molality.

Solution:

$15\text{ ppm}$ means $15\text{ g}$ of $\text{CHCl}_3$ in $10^6\text{ g}$ of solution.

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

Moles of $\text{CHCl}_3$ ($M=119.5\text{ g/mol}$) $= \frac{15}{119.5} = 0.1255\text{ mol}$.

Mass of solvent $\approx 10^6\text{ g} = 1000\text{ kg}$.

Molality ($m$) $= \frac{\text{Moles}}{\text{Mass of solvent (kg)}} = \frac{0.1255}{1000} = \mathbf{1.255 \times 10^{-4}\text{ m}}$.

Q6. Density & Molarity JEE Mains

Concentrated nitric acid used in laboratory is $68\%$ nitric acid by mass in aqueous solution. What should be the molarity of the solution if the density is $1.504\text{ g/mL}$?

Solution:

Formula: $M = \frac{\%\text{ w/w} \times \text{density} \times 10}{\text{Molar Mass}}$

Molar mass of $\text{HNO}_3 = 1 + 14 + 48 = 63\text{ g/mol}$.

$M = \frac{68 \times 1.504 \times 10}{63} = \frac{1022.72}{63} = \mathbf{16.23\text{ M}}$.

TOPIC 2: Vapour Pressure & Raoult's Law

Q7. Liquid-Liquid Solutions CBSE

Vapour pressure of chloroform ($\text{CHCl}_3$) and dichloromethane ($\text{CH}_2\text{Cl}_2$) at $298\text{ K}$ are $200\text{ mm Hg}$ and $415\text{ mm Hg}$ respectively. Calculate the vapour pressure of the solution prepared by mixing $25.5\text{ g}$ of $\text{CHCl}_3$ and $40\text{ g}$ of $\text{CH}_2\text{Cl}_2$ at $298\text{ K}$.

Solution:

$M(\text{CHCl}_3) = 119.5\text{ g/mol}$. Moles $n_1 = \frac{25.5}{119.5} = 0.213\text{ mol}$.

$M(\text{CH}_2\text{Cl}_2) = 85\text{ g/mol}$. Moles $n_2 = \frac{40}{85} = 0.47\text{ mol}$.

Total moles $= 0.213 + 0.47 = 0.683\text{ mol}$.

$x_1 = \frac{0.213}{0.683} = 0.312$. $x_2 = 1 - 0.312 = 0.688$.

$P_{\text{total}} = P^\circ_1 x_1 + P^\circ_2 x_2$

$P_{\text{total}} = (200 \times 0.312) + (415 \times 0.688) = 62.4 + 285.5 = \mathbf{347.9\text{ mm Hg}}$.

Q8. Vapour Phase Composition JEE Mains

For the solution in Q7, calculate the mole fraction of each component in the vapour phase.

Solution:

Using Dalton's Law: $P_i = y_i \times P_{\text{total}}$ (where $y_i$ is mole fraction in vapour phase).

From Q7: $P_1(\text{CHCl}_3) = 62.4\text{ mm Hg}$, $P_2(\text{CH}_2\text{Cl}_2) = 285.5\text{ mm Hg}$, and $P_{\text{total}} = 347.9\text{ mm Hg}$.

$y_1(\text{CHCl}_3) = \frac{P_1}{P_{\text{total}}} = \frac{62.4}{347.9} = \mathbf{0.18}$.

$y_2(\text{CH}_2\text{Cl}_2) = \frac{P_2}{P_{\text{total}}} = \frac{285.5}{347.9} = \mathbf{0.82}$.

(Note: The vapour phase is richer in the more volatile component, $\text{CH}_2\text{Cl}_2$).

Q9. Relative Lowering of VP CBSE

The vapour pressure of pure benzene at a certain temperature is $0.850\text{ bar}$. A non-volatile solute weighing $0.5\text{ g}$ is added to $39.0\text{ g}$ of benzene ($M=78\text{ g/mol}$). V.P. of solution becomes $0.845\text{ bar}$. What is the molar mass of solute?

Solution:

RLVP formula: $\frac{P^\circ - P_{\text{s}}}{P^\circ} = \frac{w_2 \times M_1}{M_2 \times w_1}$

$\frac{0.850 - 0.845}{0.850} = \frac{0.5 \times 78}{M_2 \times 39}$

$\frac{0.005}{0.850} = \frac{39}{39 \times M_2} \implies 0.00588 = \frac{1}{M_2}$

$M_2 = \frac{1}{0.00588} = \mathbf{170\text{ g/mol}}$.

Q10. RLVP exact formula JEE Mains

V.P. of water at $293\text{ K}$ is $17.535\text{ mm Hg}$. Calculate V.P. of water at $293\text{ K}$ when $25\text{ g}$ of glucose is dissolved in $450\text{ g}$ of water.

Solution:

Moles of glucose ($n_2$) $= \frac{25}{180} = 0.139\text{ mol}$.

Moles of water ($n_1$) $= \frac{450}{18} = 25\text{ mol}$.

Mole fraction ($x_2$) $= \frac{0.139}{25 + 0.139} = 0.0055$.

$\frac{P^\circ - P_{\text{s}}}{P^\circ} = x_2 \implies \frac{17.535 - P_{\text{s}}}{17.535} = 0.0055$

$17.535 - P_{\text{s}} = 0.096 \implies P_{\text{s}} = \mathbf{17.439\text{ mm Hg}}$.

Q11. Mass from RLVP CBSE

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

Solution:

Let initial V.P. be $P^\circ$. Final V.P. ($P_{\text{s}}$) $= 0.8 P^\circ$.

Molar mass of octane ($\text{C}_8\text{H}_{18}$) $= 114\text{ g/mol}$.

Moles of solvent ($n_1$) $= \frac{114}{114} = 1\text{ mol}$.

$\frac{P^\circ - 0.8P^\circ}{P^\circ} = \frac{n_2}{n_1 + n_2} \implies 0.2 = \frac{n_2}{1 + n_2}$

0.2 + 0.2$n_2$ $= n_2 \implies 0.8n_2 = 0.2 \implies n_2 = 0.25\text{ mol}$.

$\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.25 \times 40 = \mathbf{10\text{ g}}$.

Q12. Raoult's Law Graph JEE Mains

Two liquids X and Y form an ideal solution. At $300\text{ K}$, vapour pressure of solution containing $1\text{ mol}$ of X and $3\text{ mol}$ of Y is $550\text{ mmHg}$. At same temp, if $1\text{ mol}$ of Y is further added, VP increases by $10\text{ mmHg}$. Find $P^\circ_X$ and $P^\circ_Y$.

Solution:

Case 1: $x_X = \frac{1}{4}$, $x_Y = \frac{3}{4}$. $P_{\text{total}} = 550\text{ mmHg}$.

$P^\circ_X\left(\frac{1}{4}\right) + P^\circ_Y\left(\frac{3}{4}\right) = 550 \implies P^\circ_X + 3P^\circ_Y = 2200 \quad \text{(Eq. 1)}$

Case 2: $1\text{ mol}$ Y added. Total moles $= 5$. $x_X = \frac{1}{5}$, $x_Y = \frac{4}{5}$. $P_{\text{total}} = 560\text{ mmHg}$.

$P^\circ_X\left(\frac{1}{5}\right) + P^\circ_Y\left(\frac{4}{5}\right) = 560 \implies P^\circ_X + 4P^\circ_Y = 2800 \quad \text{(Eq. 2)}$

Subtract Eq. 1 from Eq. 2: $\mathbf{P^\circ_Y = 600\text{ mm Hg}}$.

Put in Eq. 1: $P^\circ_X + 3(600) = 2200 \implies \mathbf{P^\circ_X = 400\text{ mm Hg}}$.

TOPIC 3: Elevation of BP & Depression of FP

Q13. Boiling Point Elevation CBSE

Boiling point of water at $750\text{ mm Hg}$ is $99.63^\circ\text{C}$. How much sucrose is to be added to $500\text{ g}$ of water such that it boils at $100^\circ\text{C}$? ($K_b \text{ for water} = 0.52\text{ K kg mol}^{-1}$)

Solution:

$\Delta T_b = 100 - 99.63 = 0.37^\circ\text{C} = 0.37\text{ K}$.

Mass of water ($w_1$) $= 500\text{ g} = 0.5\text{ kg}$.

Molar mass sucrose ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$) $= 342\text{ g/mol}$.

$\Delta T_b = K_b \times \frac{w_2}{M_2 \times w_1} \implies 0.37 = 0.52 \times \frac{w_2}{342 \times 0.5}$

$w_2 = \frac{0.37 \times 342 \times 0.5}{0.52} = \mathbf{121.67\text{ g}}$.

Q14. Freezing Point Depression CBSE

$45\text{ g}$ of ethylene glycol ($\text{C}_2\text{H}_6\text{O}_2$) is mixed with $600\text{ g}$ of water. Calculate (a) freezing point depression and (b) freezing point of solution. ($K_f = 1.86\text{ K kg mol}^{-1}$)

Solution:

Molar mass of ethylene glycol $= 62\text{ g/mol}$.

Moles of solute $= \frac{45}{62} = 0.725\text{ mol}$.

Mass of solvent $= 0.6\text{ kg}$.

Molality ($m$) $= \frac{0.725}{0.6} = 1.2\text{ m}$.

(a) $\Delta T_f = K_f \times m = 1.86 \times 1.2 = \mathbf{2.2\text{ K}}$.

(b) $T_f(\text{solution}) = T^\circ_f - \Delta T_f = 273.15 - 2.2 = \mathbf{270.95\text{ K}}$ (or $-2.2^\circ\text{C}$).

Q15. Ice Separation JEE Mains

A solution containing $34.2\text{ g}$ of cane sugar dissolved in $500\text{ g}$ of water is cooled to $-0.372^\circ\text{C}$. What mass of ice will separate out? ($K_f = 1.86\text{ K kg/mol}$)

Solution:

At $-0.372^\circ\text{C}$, solution is in equilibrium with ice. $\Delta T_f = 0.372\text{ K}$.

Moles of sugar $= \frac{34.2}{342} = 0.1\text{ mol}$.

Let $W\text{ kg}$ be the mass of water remaining in liquid state.

$\Delta T_f = K_f \times \frac{\text{Moles}}{W} \implies 0.372 = 1.86 \times \frac{0.1}{W}$

$W = \frac{1.86 \times 0.1}{0.372} = 0.5\text{ kg} = 500\text{ g}$.

Since initially water was $500\text{ g}$, the remaining liquid water is still $500\text{ g}$, so **$0\text{ g}$ of ice separates** (it just starts freezing).

Q16. K_b / K_f Ratio JEE Mains

For a dilute aqueous solution, the freezing point is $-0.186^\circ\text{C}$. Calculate the boiling point of this solution. ($K_f = 1.86\text{ K kg mol}^{-1}$, $K_b = 0.512\text{ K kg mol}^{-1}$)

Solution:

Since both processes use the same solution, molality ($m$) is constant.

$\Delta T_f = K_f \times m \implies 0.186 = 1.86 \times m \implies m = 0.1\text{ mol/kg}$.

$\Delta T_b = K_b \times m = 0.512 \times 0.1 = 0.0512^\circ\text{C}$.

Boiling point of solution $= T^\circ_b + \Delta T_b = 100 + 0.0512 = \mathbf{100.0512^\circ\text{C}}$.

Q17. Mass from ΔT_b CBSE

$18\text{ g}$ of glucose is dissolved in $1\text{ kg}$ of water in a saucepan. At what temperature will water boil at $1.013\text{ bar}$? ($K_b \text{ for water} = 0.52\text{ K kg/mol}$)

Solution:

Molar mass of glucose ($M_2$) $= 180\text{ g/mol}$.

Molality ($m$) $= \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{18 / 180}{1} = 0.1\text{ mol/kg}$.

$\Delta T_b = K_b \times m = 0.52 \times 0.1 = 0.052\text{ K}$.

Boiling Point of Solution $= 373.15 + 0.052 = \mathbf{373.202\text{ K}}$ (or $100.052^\circ\text{C}$).

Q18. Calculating Molar Mass JEE Mains

Addition of $0.643\text{ g}$ of a compound to $50\text{ mL}$ of benzene (density: $0.879\text{ g/mL}$) lowers the freezing point from $5.51^\circ\text{C}$ to $5.03^\circ\text{C}$. If $K_f$ for benzene is $5.12\text{ K kg/mol}$, calculate molar mass of the compound.

Solution:

$\Delta T_f = 5.51 - 5.03 = 0.48\text{ K}$.

Mass of solvent (benzene) $= V \times d = 50 \times 0.879 = 43.95\text{ g} = 0.04395\text{ kg}$.

$\Delta T_f = \frac{K_f \times w_2}{M_2 \times w_1} \implies 0.48 = \frac{5.12 \times 0.643}{M_2 \times 0.04395}$

$M_2 = \frac{5.12 \times 0.643}{0.48 \times 0.04395} = \mathbf{156.06\text{ g/mol}}$.

TOPIC 4: Osmotic Pressure & van't Hoff Factor

Q19. Osmotic Pressure CBSE

$200\text{ cm}^3$ of an aqueous solution of a protein contains $1.26\text{ g}$ of the protein. The osmotic pressure of such a solution at $300\text{ K}$ is found to be $2.57 \times 10^{-3}\text{ bar}$. Calculate the molar mass of the protein.

Solution:

$\pi = CRT = \frac{w_2 RT}{M_2 V}$

$\pi = 2.57 \times 10^{-3}\text{ bar}$, $V = 0.2\text{ L}$, $T = 300\text{ K}$.

$R = 0.083\text{ L bar K}^{-1}\text{ mol}^{-1}$.

$M_2 = \frac{w_2 RT}{\pi V} = \frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.2} = \frac{31.374}{0.000514} = \mathbf{61,039\text{ g/mol}}$.

Q20. Isotonic Solutions CBSE

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

Solution:

For cane sugar ($M = 342\text{ g/mol}$): $\Delta T_f = 273.15 - 271 = 2.15\text{ K}$.

$\text{Molality} = \frac{5 / 342}{0.095\text{ kg}} = 0.154\text{ m}$.

$K_f = \frac{2.15}{0.154} = 13.96\text{ K kg mol}^{-1}$.

For glucose ($M = 180\text{ g/mol}$): $5\%$ means $5\text{ g}$ glucose in $95\text{ g}$ water.

$\text{Molality} = \frac{5 / 180}{0.095\text{ kg}} = 0.292\text{ m}$.

$\Delta T_f(\text{glucose}) = K_f \times m = 13.96 \times 0.292 = 4.08\text{ K}$.

Freezing point $= 273.15 - 4.08 = \mathbf{269.07\text{ K}}$.

Q21. Dissociation (i > 1) JEE Mains

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

Solution:

$\pi = i \times \frac{n}{V} \times RT$

$0.75 = 2.47 \times \left(\frac{n}{2.5}\right) \times 0.0821 \times 300 \implies 0.75 = n \times 24.33$

$n = \frac{0.75}{24.33} = 0.0308\text{ mol}$.

Molar mass of $\text{CaCl}_2 = 111\text{ g/mol}$.

$\text{Mass} = n \times M = 0.0308 \times 111 = \mathbf{3.42\text{ g}}$.

Q22. Association (i < 1) JEE Mains

$2\text{ g}$ of benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$) dissolved in $25\text{ g}$ of benzene shows a depression in freezing point equal to $1.62\text{ K}$. Molal depression constant for benzene is $4.9\text{ K kg mol}^{-1}$. What is the percentage association of acid if it forms dimer?

Solution:

Observed $\Delta T_f = 1.62\text{ K}$.

Calculated $\Delta T_f = K_f \times \frac{w_2}{M_2 \times w_1} = 4.9 \times \frac{2}{122 \times 0.025} = 3.21\text{ K}$.

$i = \frac{\text{Observed}}{\text{Calculated}} = \frac{1.62}{3.21} = 0.504$.

Dimerization: $2\text{A} \rightleftharpoons \text{A}_2$. Therefore, $i = 1 - \frac{\alpha}{2}$.

$0.504 = 1 - \frac{\alpha}{2} \implies \frac{\alpha}{2} = 0.496 \implies \alpha = 0.992$.

$\text{Percentage association} = \mathbf{99.2\%}$.

Q23. Dissociation Constant JEE Mains

$0.6\text{ mL}$ of acetic acid ($\text{CH}_3\text{COOH}$, density $1.06\text{ g/mL}$) is dissolved in $1\text{ L}$ of water. The depression in freezing point observed is $0.0205^\circ\text{C}$. Calculate the van't Hoff factor. ($K_f = 1.86\text{ K kg mol}^{-1}$)

Solution:

Mass of acid $= V \times d = 0.6\text{ mL} \times 1.06\text{ g/mL} = 0.636\text{ g}$.

Moles $= \frac{0.636}{60\text{ g/mol}} = 0.0106\text{ mol}$.

Molality ($m$) $= \frac{0.0106\text{ mol}}{1\text{ kg}} = 0.0106\text{ m}$.

Calculated $\Delta T_f = 1.86 \times 0.0106 = 0.0197\text{ K}$.

$i = \frac{\text{Observed }\Delta T_f}{\text{Calculated }\Delta T_f} = \frac{0.0205}{0.0197} = \mathbf{1.041}$.

(Here, $\alpha = i - 1 = 0.041$, indicating $4.1\%$ weak dissociation).

Q24. Elevation with i CBSE

Calculate the boiling point of a solution prepared by adding $15.00\text{ g}$ of $\text{NaCl}$ to $250\text{ g}$ of water. ($K_b = 0.512\text{ K kg mol}^{-1}$, Molar mass of $\text{NaCl} = 58.44\text{ g/mol}$)

Solution:

Assume $100\%$ dissociation for strong electrolyte $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$, so $i = 2$.

Moles of $\text{NaCl} = \frac{15.00}{58.44} = 0.256\text{ mol}$.

Molality ($m$) $= \frac{0.256}{0.250\text{ kg}} = 1.024\text{ m}$.

$\Delta T_b = i \times K_b \times m = 2 \times 0.512 \times 1.024 = 1.048\text{ K}$.

Boiling point $= 373.15 + 1.048 = \mathbf{374.198\text{ K}}$ (or $101.05^\circ\text{C}$).