Q.Calculate the mass of ascorbic acid (Vitamin C, $C_6H_8O_6$) 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$^{-1}$.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.A solution containing 7.5 g of urea (molar mass = 60 g mol$^{-1}$) in 1 kg of water freezes at the same temperature as another solution containing 15 g of solute X, in the same amount of water. The molar mass of X (g mol$^{-1}$) is (A) 60 (B) 180 (C) 120 (D) 240

   ›Reveal solutionSolution

   This tests that identical freezing-point depression in the same solvent mass implies identical molality, letting you back-calculate an unknown solute's molar mass.

   Concept and Intuition

   Freezing point depression is a colligative property: $\Delta T_f = K_f \times m$ (for a non-electrolyte solute), depending only on the molality of solute particles, not their identity. If two solutions in the same mass of the same solvent freeze at the same temperature, they must have the same molality (assuming neither dissociates/associates, as is the case for urea and a general non-electrolyte X here).

   Step-by-Step Solution

   1. Moles of urea $= \dfrac{7.5\ \mathrm{g}}{60\ \mathrm{g\,mol^{-1}}} = 0.125$ mol.
   2. Molality of urea solution $= \dfrac{0.125\ \mathrm{mol}}{1\ \mathrm{kg}} = 0.125\ m$.
   3. Since solution X has the same freezing point (same solvent mass, 1 kg water), it must have the same molality: $0.125\ m$.
   4. So moles of X in 1 kg water $= 0.125$ mol. …

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2. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.At T(K), the vapour pressure of an aqueous solution of a non-volatile solute, whose mole fraction is 0.02 is found to be 34.65 mm Hg. What is the vapour pressure (in mm Hg) of pure water at the same temperature ? (A) 35.70 (B) 35.36 (C) 35.00 (D) 34.30

   ›Reveal solutionSolution

   Using Raoult's law for a non-volatile solute, the relative lowering of vapour pressure equals the solute mole fraction, giving the pure solvent's vapour pressure as 35.36 mmHg.

   Concept and Intuition

   For a solution of a non-volatile solute in a volatile solvent, Raoult's law gives the relative lowering of vapour pressure as equal to the mole fraction of the solute: $\dfrac{P^0 - P}{P^0} = x_{\text{solute}}$, where $P^0$ is the vapour pressure of the pure solvent and $P$ is that of the solution.

   Step-by-Step Solution

   1. Given: solute mole fraction $x_{\text{solute}} = 0.02$, solution vapour pressure $P = 34.65$ mmHg.
   2. Apply Raoult's law: $\dfrac{P^0 - 34.65}{P^0} = 0.02$.
   3. Rearranging: $P^0 - 34.65 = 0.02\,P^0 \Rightarrow P^0(1 - 0.02) = 34.65 \Rightarrow 0.98\,P^0 = 34.65$. …

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3. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.Benzoic acid molecules undergo dimerisation in benzene. 2.44 g of benzoic acid when dissolved in 30 g of benzene caused depression in freezing point of 2 K. What is the percentage of association of it? (Given $K_f(\mathrm{C_6H_6}) = 5\ \mathrm{K\,kg\,mol^{-1}}$; molar mass of benzoic acid $= 122\ \mathrm{g\,mol^{-1}}$) (A) 80 (B) 70 (C) 60 (D) 90

   ›Reveal solutionSolution

   This tests the van't Hoff factor for association (dimerisation) using freezing-point depression data. The answer is 80% association.

   Concept and Intuition

   When solute molecules associate (like benzoic acid dimerising in benzene via hydrogen bonding), the effective number of particles in solution decreases, which lowers the observed colligative effect below the value predicted by the simple formula. The van't Hoff factor $i$ quantifies this: $i = \dfrac{\text{observed colligative property}}{\text{calculated (no association) colligative property}}$, and for a solute that dimerises with degree of association $\alpha$, $i = 1 - \dfrac{\alpha}{2}$ (since two molecules become one particle for every $\alpha$ fraction that associates).

   Step-by-Step Solution

   1. Moles of benzoic acid $=\dfrac{2.44}{122}=0.02$ mol.
   2. Molality $m=\dfrac{0.02\ \text{mol}}{0.030\ \text{kg}}=0.6667\ \mathrm{mol\,kg^{-1}}$.
   3. Calculated (no association) $\Delta T_f = K_f m = 5\times0.6667=3.333$ K. …

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4. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.At T(K), the vapour pressure of water is $x$ kPa. What is the vapour pressure (in kPa) of 1 molal solution containing non-volatile solute? (A) $1.018\,x$ (B) $0.8\,x$ (C) $0.972\,x$ (D) $0.982\,x$

   ›Reveal solutionSolution

   Uses Raoult's law: the relative lowering of vapour pressure equals the mole fraction of the non-volatile solute, computed from the given molality. Answer: (D).

   Concept and Intuition

   Raoult's law for a solution of a non-volatile solute in a volatile solvent states that the relative lowering of the solvent's vapour pressure equals the mole fraction of the solute: $\dfrac{p^0-p}{p^0}=x_{\text{solute}}$. A '1 molal' solution means 1 mole of solute is dissolved in exactly 1 kg (1000 g) of solvent, which lets us directly compute the mole fraction of solute from the known molar mass of water.

   Step-by-Step Solution

   1. Molality $=1$ mol solute per 1000 g (1 kg) water.
   2. Moles of water in 1000 g $=\dfrac{1000}{18}=55.56$ mol.
   3. Mole fraction of solute $x_2=\dfrac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{water}}}=\dfrac{1}{1+55.56}=\dfrac{1}{56.56}=0.01768$.
   4. By Raoult's law, $\dfrac{p^0-p}{p^0}=x_2=0.01768$. …

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5. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.Elements X and Y form two non-volatile compounds (XY and $\mathrm{XY_3}$). When 10 g of XY is dissolved in 50 g of ethanol, the depression in freezing point ($\Delta T_f$) was 5.333 K. When 10 g of $\mathrm{XY_3}$ is dissolved in 50 g of ethanol, the was ($\Delta T_f$) 2.2857 K. What are the atomic weights of X and Y respectively? ($K_f = 2\ \mathrm{K\ kg\ mol^{-1}}$) (A) 50 u, 50 u (B) 25 u, 25 u (C) 75 u, 100 u (D) 25 u, 50 u

   ›Reveal solutionSolution

   Depression in freezing point gives the molar mass of each compound (via $\Delta T_f=K_f m$); two simultaneous equations in atomic weights of X and Y then solve directly. Answer: (D), X = 25 u, Y = 50 u.

   Concept and Intuition

   Depression of freezing point is a colligative property: $\Delta T_f=K_f\times m$, where $m$ is the molality of the solute. Knowing the mass of solute dissolved and the mass of solvent lets us back-calculate the molar mass of the solute from the measured $\Delta T_f$. Doing this for both compounds XY and $\mathrm{XY_3}$ gives two independent linear equations in the atomic weights of X and Y, which can be solved simultaneously.

   Step-by-Step Solution

   1. For XY: molality $m_1=\dfrac{\Delta T_f}{K_f}=\dfrac{5.333}{2}=2.6665\ \mathrm{mol\,kg^{-1}}$.
   2. Moles of XY dissolved in 50 g $(0.05\ \text{kg})$ ethanol: $n_1=m_1\times 0.05=0.13333$ mol.
   3. Molar mass of XY $=\dfrac{\text{mass}}{n_1}=\dfrac{10}{0.13333}\approx 75.0\ \text{g/mol}$. So $M_X+M_Y=75$.
   4. For $\mathrm{XY_3}$: molality $m_2=\dfrac{2.2857}{2}=1.14285\ \mathrm{mol\,kg^{-1}}$.
   5. Moles of $\mathrm{XY_3}$ in 50 g ethanol: $n_2=1.14285\times0.05=0.057143$ mol. …

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6. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.Ethylene glycol ($\mathrm{C_2H_6O_2}$) is used as an antifreeze in cars. In a certain place, it is desired that water freezes at 258K. How much weight of ethylene glycol is needed to prevent separation of ice from 500 g of water? ($K_f$ of water = 1.86 K kg mol$^{-1}$) (A) 125 g (B) 62.5 g (C) 250 g (D) 175 g

   ›Reveal solutionSolution

   The required freezing-point depression (15 K) fixes the molality via $\Delta T_f = K_f m$; converting that molality to moles in 500 g of water and then to mass gives exactly 250 g of ethylene glycol.

   Concept and Intuition

   Freezing-point depression is a colligative property: $\Delta T_f = K_f \times m$, where $m$ is the molality of solute particles (ethylene glycol does not dissociate, so its molality directly gives the depression). Once molality is known, multiplying by the mass of solvent (in kg) gives moles of solute, and multiplying by molar mass gives the required mass.

   Step-by-Step Solution

   1. Normal freezing point of water = 273 K; desired freezing point = 258 K, so $\Delta T_f = 273-258 = 15$ K.
   2. $\Delta T_f = K_f\, m \Rightarrow m = \dfrac{15}{1.86} = 8.065$ mol/kg.
   3. Mass of water = 500 g = 0.5 kg, so moles of ethylene glycol needed $= 8.065 \times 0.5 = 4.032$ mol. …

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7. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.x g of benzoic acid (molar mass $= 122\ \mathrm{g\,mol^{-1}}$) is dissolved in 50 g of benzene. Its freezing point was found to be 277.82 K. What is the value of 'x'? (Given: $K_f$ of benzene $=5.1\ \mathrm{K\,kg\,mol^{-1}}$, freezing point of benzene $= 278.45$ K and Van't Hoff's factor of benzoic acid $= 0.5$) (A) 0.5 (B) 1.5 (C) 0.75 (D) 1.0

   ›Reveal solutionSolution

   Using the freezing-point depression formula with the given van't Hoff factor (which accounts for benzoic acid's dimerization in benzene) gives $x = 1.5$ g.

   Concept and Intuition

   Benzoic acid dimerizes in benzene (via hydrogen bonding), so its effective van't Hoff factor is less than 1 (given as 0.5, i.e. near-complete dimerization); the depression in freezing point is scaled by this factor: $\Delta T_f = i \cdot K_f \cdot m$.

   Step-by-Step Solution

   1. $\Delta T_f = 278.45 - 277.82 = 0.63$ K.
   2. $\Delta T_f = i K_f m \Rightarrow m = \dfrac{\Delta T_f}{i K_f} = \dfrac{0.63}{0.5 \times 5.1} = \dfrac{0.63}{2.55} = 0.2471\ \mathrm{mol/kg}$.
   3. Molality $m = \dfrac{x/122}{0.050\ \text{kg}}$ (since 50 g $= 0.050$ kg of benzene solvent).
   4. So $\dfrac{x}{122} = 0.2471 \times 0.050 = 0.012353\ \text{mol}$. …

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8. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.1 g of $\mathrm{XY_2}$ is dissolved in 20 g of $\mathrm{C_6H_6}$. The $\Delta T_f$ of resultant solution is 2.318 K. When 1 g of $\mathrm{XY_4}$ is dissolved in 20 g of $\mathrm{C_6H_6}$, its $\Delta T_f$ is found to be 1.314 K. What are the atomic masses of X and Y respectively? ($K_f$ of $\mathrm{C_6H_6}$ is 5.1 K kg mol$^{-1}$) (A) 42 u, 26 u (B) 38 u, 30 u (C) 30 u, 38 u (D) 26 u, 42 u

   ›Reveal solutionSolution

   Freezing-point depression gives the molar masses of $\mathrm{XY_2}$ (≈110) and

   $\mathrm{XY_4}$ (≈194); solving $X+2Y=110$ and $X+4Y=194$ simultaneously gives

   $X=26$ u and $Y=42$ u.

   Concept and Intuition

   The depression in freezing point is $\Delta T_f = K_f \times m$, where molality

   $m = \dfrac{(w_2/M)}{W_1(\text{kg})}$, $w_2$ = mass of solute, $M$ = its molar mass,

   $W_1$ = mass of solvent in kg. Rearranging, $M = \dfrac{K_f \times w_2}{\Delta T_f \times W_1(\text{kg})}$.

   Since both compounds share the elements X and Y, writing two equations in the unknown

   atomic masses lets us solve them simultaneously — a standard "back out the atomic mass"

   colligative-properties problem.

   Step-by-Step Solution

   1. For $\mathrm{XY_2}$: $w_2=1$ g, $W_1=20$ g $=0.020$ kg, $K_f=5.1$, $\Delta T_f=2.318$ K.

   $M_1=\dfrac{5.1\times1}{2.318\times0.020}=\dfrac{5.1}{0.04636}\approx110\ \text{g/mol}$

   2. For $\mathrm{XY_4}$: $\Delta T_f=1.314$ K.

   $M_2=\dfrac{5.1\times1}{1.314\times0.020}=\dfrac{5.1}{0.02628}\approx194\ \text{g/mol}$

   3. Let atomic masses be $X$ and $Y$. Then: $X+2Y=110 \qquad (i)$ …

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9. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.Benzoic acid undergoes dimerization in benzene. $x$ g of benzoic acid (molar mass 122 g mol$^{-1}$) is dissolved in 49 g of benzene. The depression in freezing point is 1.12 K. If degree of association of acid is 88 %, what is the value of $x$? ($K_f$ for benzene = 4.9 K kg mol$^{-1}$) (A) 2.44 (B) 1.22 (C) 3.66 (D) 4.88

   ›Reveal solutionSolution

   Benzoic acid partially dimerizes in benzene; using the van't Hoff factor for association ($i = 1-\alpha+\alpha/n$) in the freezing-point-depression formula gives $x = 2.44$ g.

   Concept and Intuition

   When a solute associates (here, benzoic acid dimerizes via H-bonding in a non-polar solvent like benzene), the effective number of particles in solution is LESS than the number of solute formula units dissolved, so the van't Hoff factor $i < 1$. For association into $n$-mers with degree of association $\alpha$: $i = 1 - \alpha + \dfrac{\alpha}{n}$. This $i$ then scales the ideal colligative-property formula, here freezing point depression $\Delta T_f = i\,K_f\,m$.

   Step-by-Step Solution

   1. Degree of association $\alpha = 0.88$, dimerization means $n=2$.
   2. Van't Hoff factor: $i = 1 - \alpha + \dfrac{\alpha}{n} = 1 - 0.88 + \dfrac{0.88}{2} = 0.12 + 0.44 = 0.56$.
   3. Molality: $m = \dfrac{x/122}{0.049\ \text{kg}}$ (x g of solute, molar mass 122 g/mol, in 49 g = 0.049 kg benzene).
   4. Freezing point depression: $\Delta T_f = i\,K_f\,m \Rightarrow 1.12 = 0.56 \times 4.9 \times \dfrac{x/122}{0.049}$. …

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10. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
    Q.Match the following List – I | List – II A. $\pi$ | I. $iK_bm$ B. $\Delta T_f$ | II. $iK_fm$ C. $\Delta T_b$ | III. $iCST$ D. $\dfrac{P^0-P_s}{P^0}$ | IV. $i\left(\dfrac{n_2}{n_1+in_2}\right)$ The correct answer is (A) A – III, B – II, C – I, D – IV (B) A – IV, B – III, C – II, D – I (C) A – I, B – III, C – IV, D – II (D) A – III, B – IV, C – II, D – I

    ›Reveal solutionSolution

    Straightforward matching of the four colligative property expressions (each modified with the van't Hoff factor $i$) to their formulas.

    Concept and Intuition

    All four colligative properties get an extra factor $i$ (the van't Hoff factor) when the solute dissociates or associates in solution, since colligative properties depend on the total number of solute particles, not just formula units.

    Step-by-Step Solution

    1. A. Osmotic pressure $\pi = iCST$ (van't Hoff equation, analogous to ideal gas law) $\rightarrow$ III.
    2. B. Freezing point depression $\Delta T_f = iK_f m$ $\rightarrow$ II.
    3. C. Boiling point elevation $\Delta T_b = iK_b m$ $\rightarrow$ I.
    4. D. Relative lowering of vapor pressure $\dfrac{P^0-P_s}{P^0} = i\left(\dfrac{n_2}{n_1+in_2}\right)$ (mole fraction of solute, with total particles accounting for $i$) $\rightarrow$ IV. …

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11. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.0.5g of non-volatile solute is added to the 39g of Benzene. Then vapour pressure of solution is (vapour pressure of pure Benzene is 0.850 torr) (A) 1 : 1 (B) 1 : 12 (C) 1 : 3 (D) 1 : 6

    ›Reveal solutionSolution

    For a non-volatile solute the vapour pressure of the solution comes only from the solvent (Raoult's law). The printed options are ratios, so the intended answer per the official key is (B) 1 : 12.

    For a non-volatile solute in benzene, only benzene contributes to the vapour above the solution. By Raoult's law the solvent's partial pressure is set by its mole fraction:

    $p_{\text{soln}} = x_{\text{benzene}}\, p^{0}_{\text{benzene}}, \qquad \frac{p^{0}-p_{\text{soln}}}{p^{0}} = x_{\text{solute}}$

    The data given are $p^{0}_{\text{benzene}} = 0.850\ \text{torr}$, mass of benzene $= 39\ \text{g}$ (so $n_{\text{benzene}} = 39/78 = 0.5\ \text{mol}$), and mass of solute $= 0.5\ \text{g}$. …

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12. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.0.5g of non-volatile solute is added to the 39g of Benzene. Then vapour pressure of solution is (vapour pressure of pure Benzene is 0.850 torr) (A) 0.845 torr (B) 0.850 torr (C) 0.860 torr (D) 0.870 torr

    ›Reveal solutionSolution

    Dissolving any non-volatile solute must lower a solvent's vapour pressure below its pure value, so the only physically possible answer among the choices is 0.845 torr.

    Concept and Intuition

    Raoult's law states that for an ideal solution, $P_{solution} = P^{\circ}_{solvent}\,x_{solvent}$, where $x_{solvent} < 1$ once any solute is added. Hence $P_{solution}$ must always be strictly less than $P^{\circ}_{solvent}$ for a non-volatile solute — there is no scenario in which adding solute increases or leaves unchanged the vapour pressure.

    Step-by-Step Solution

    1. Moles of benzene $= \dfrac{39\text{ g}}{78\text{ g/mol}} = 0.5$ mol (benzene, $\mathrm{C_6H_6}$, molar mass 78 g/mol).
    2. Whatever the solute's molar mass, adding 0.5 g of it introduces a positive mole fraction of solute, so $x_{benzene} < 1$.
    3. Therefore $P_{solution} = 0.850 \times x_{benzene} < 0.850$ torr.
    4. Checking the options: 0.850 (no change — impossible), 0.860 and 0.870 (increase — impossible for a non-volatile solute), leaving only 0.845 torr as physically valid. …

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