Q.1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol$^{-1}$. Find the molar mass of the solute.

Showing the 12 most recent of 21 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The value of $K_f$ (in $K\,kg\,mol^{-1}$) of a solvent (X) is four times the value of its $K_b$. 2g of a non-volatile, non-electrolytic solute $A$ is dissolved in 200 g of solvent X. The $\Delta T_b$ of resultant solution is $YK$. 4g of $A$ is dissolved in 200 g of X and the $\Delta T_f$ of resultant solution is $ZK$. What is the ratio of Y and Z? (Molar mass of A = 100 $g\,mol^{-1}$) (A) 1 : 2 (B) 1 : 4 (C) 1 : 8 (D) 1 : 16

   ›Reveal solutionSolution

   Using $\Delta T = K \times$ molality for both cases and $K_f = 4K_b$, the ratio $Y:Z$ works out to $1:8$.

   Concept and Intuition

   Both elevation of boiling point and depression of freezing point are colligative properties proportional to molality: $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$. Since the same solute A (molar mass 100) is used in the same mass of the same solvent X in both cases, only the mass of solute taken and the relevant constant ($K_b$ vs $K_f$) differ between the two scenarios.

   Step-by-Step Solution

   1. Case 1 ($\Delta T_b = Y$): moles of A $= \dfrac{2}{100} = 0.02$ mol; mass of solvent $= 0.2$ kg. Molality $m_1 = \dfrac{0.02}{0.2} = 0.1$ mol/kg.

   $Y = K_b \times 0.1$

   2. Case 2 ($\Delta T_f = Z$): moles of A $= \dfrac{4}{100} = 0.04$ mol; mass of solvent $= 0.2$ kg. Molality $m_2 = \dfrac{0.04}{0.2} = 0.2$ mol/kg. $Z = K_f \times 0.2$ …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.An aqueous solution of a non-volatile and non-electrolytic solute boils at $100.5°C$. What will be the freezing point of the same solution? (Given $K_b = 0.512\ K\ kg\ mol^{-1}$ and $K_f = 1.86\ K\ kg\ mol^{-1}$) (A) $-2.816\ °C$ (B) $-1.816\ °C$ (C) $-0.908\ °C$ (D) $-3.632\ °C$

   ›Reveal solutionSolution

   This tests linking boiling-point elevation to freezing-point depression via a common molality; the freezing point works out to $-1.816°C$.

   Concept and Intuition

   For a dilute solution of a non-volatile, non-electrolyte solute, both the elevation in boiling point and the depression in freezing point are colligative properties proportional to the same molality of the solute: $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$. Given one property, we can back out the molality and then predict the other.

   Step-by-Step Solution

   1. Elevation in boiling point: $\Delta T_b = 100.5°C - 100°C = 0.5\ K$.
   2. Find molality from $\Delta T_b = K_b m$: $m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.5}{0.512} \approx 0.9766\ mol\,kg^{-1}$.
   3. Find depression in freezing point: $\Delta T_f = K_f m = 1.86 \times 0.9766 \approx 1.8164\ K$. …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The boiling point of 1M aqueous solution of $KCl$ (85% dissociation) having density $1.04\ g\,mL^{-1}$ is (Given: $K_b(H_2O) = 0.52\ K\,kg\,mol^{-1}$, molar mass of $KCl = 74.5\ g\,mol^{-1}$) (A) $100.096°C$ (B) $100.996°C$ (C) $100.896°C$ (D) $100.796°C$

   ›Reveal solutionSolution

   This tests boiling-point elevation with a dissociating electrolyte, requiring first converting molarity to molality via the given density. The boiling point comes out to 100.996 °C.

   Concept and Intuition

   Boiling point elevation depends on molality (not molarity), and for an electrolyte that partially dissociates, the van't Hoff factor $i$ accounts for the actual number of particles in solution.

   Step-by-Step Solution

   1. Take 1 litre (1000 mL) of the 1M KCl solution as the basis. Moles of KCl $= 1$ mol.
   2. Mass of the whole solution $= \text{density} \times \text{volume} = 1.04\ g/mL \times 1000\ mL = 1040\ g$.
   3. Mass of solute (KCl) $= 1\ mol \times 74.5\ g/mol = 74.5\ g$.
   4. Mass of solvent (water) $= 1040 - 74.5 = 965.5\ g = 0.9655\ kg$.
   5. Molality $m = \dfrac{\text{moles solute}}{\text{kg solvent}} = \dfrac{1}{0.9655} \approx 1.0357\ mol\,kg^{-1}$.
   6. KCl dissociates as $KCl \to K^+ + Cl^-$, so $n = 2$ ions per formula unit. With degree of dissociation $\alpha = 0.85$: van't Hoff factor $i = 1 + \alpha(n-1) = 1 + 0.85(2-1) = 1.85$. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.At 300 K, $x$ moles of $CaCl_2$ ($i = 2.5$ ; molar mass $= 111\ g\ mol^{-1}$) is dissolved in 2.5 L of water. The osmotic pressure of resultant solution is 0.75 atm. What is $\Delta T_b$ of solution? (density of water $= 1\ g\ mL^{-1}$ ; $K_b = 0.52\ K\ kg\ mol^{-1}$ ; $R = 0.08\ L\ atm\ mol^{-1}K^{-1}$) (A) 0.016 K (B) 0.032 K (C) 0.048 K (D) 0.064 K

   ›Reveal solutionSolution

   Using the osmotic-pressure data to find the molar concentration, then the molality, and finally applying $\Delta T_b = iK_bm$ gives $\Delta T_b \approx 0.016\ K$.

   Concept and Intuition

   Both osmotic pressure and boiling-point elevation are colligative properties that depend on the effective number of particles in solution (captured by the van't Hoff factor $i$). We first use the osmotic pressure relation $\pi = iCRT$ to back out the molar concentration $C$ of $CaCl_2$, convert that to molality using the mass of water given, and then apply the boiling-point elevation formula.

   Step-by-Step Solution

   1. From $\pi = iCRT$: $C = \dfrac{\pi}{iRT} = \dfrac{0.75}{2.5 \times 0.08 \times 300} = \dfrac{0.75}{60} = 0.0125\ mol\ L^{-1}$.
   2. Moles of $CaCl_2$, $x = C \times V = 0.0125 \times 2.5\ L = 0.03125\ mol$.
   3. Mass of water (solvent) $= 2.5\ L \times 1000\ g/L\ (\text{density}=1\ g/mL) = 2500\ g = 2.5\ kg$.
   4. Molality, $m = \dfrac{x}{\text{kg of solvent}} = \dfrac{0.03125}{2.5} = 0.0125\ mol\ kg^{-1}$. …

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5. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A solution of urea in water has a boiling point of 100.18 °C. What is the freezing point of the same solution, if $K_f$ and $K_b$ of water are 1.86 and 0.52 K kg mol$^{-1}$, respectively ? (Boiling point of water = 100 °C) (A) $-0.34\ ^\circ C$ (B) $-0.22\ ^\circ C$ (C) $-0.64\ ^\circ C$ (D) $-0.32\ ^\circ C$

   ›Reveal solutionSolution

   This links two colligative properties (boiling point elevation and freezing point depression) through the common molality of the solution.

   Concept and Intuition

   Both boiling-point elevation and freezing-point depression are colligative properties proportional to the same molality of solute particles: $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$. Given one, we can find the molality and then use it to get the other.

   Step-by-Step Solution

   1. Boiling point elevation: $\Delta T_b = 100.18 - 100 = 0.18\,^\circ C$.
   2. Molality from $\Delta T_b = K_b m$: $m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.18}{0.52} = 0.3462\ \text{mol/kg}$.
   3. Freezing point depression: $\Delta T_f = K_f m = 1.86 \times 0.3462 = 0.6439\,^\circ C \approx 0.64\,^\circ C$. …

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6. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.The molal depression constant of water ($K_f$) is $1.86\ \mathrm{K\,kg\,mol^{-1}}$. What is the approximate $\Delta_{fus}H$ (in $\mathrm{kJ\,mol^{-1}}$) of water if it freezes at 273 K? (A) 3 (B) 6 (C) 9 (D) 12

   ›Reveal solutionSolution

   The cryoscopic-constant formula relates $K_f$ to the solvent's molar mass, freezing point, and molar enthalpy of fusion; solving gives $\Delta_{fus}H \approx 6$ kJ/mol for water.

   Concept and Intuition

   The molal depression constant $K_f$ of a solvent is fundamentally derived from its freezing point and its enthalpy of fusion (the energy needed to melt the solid), via the thermodynamic relation:

   $K_f = \dfrac{R\,T_f^2\,M}{1000\,\Delta_{fus}H}$

   where $M$ is the solvent's molar mass (g/mol), $T_f$ its freezing point (K), and $\Delta_{fus}H$ its molar enthalpy of fusion (J/mol).

   Step-by-Step Solution

   1. Rearranging for $\Delta_{fus}H$: $\Delta_{fus}H = \dfrac{R\,T_f^2\,M}{1000\,K_f}$.
   2. Substitute $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$, $T_f=273$ K, $M=18\ \mathrm{g/mol}$, $K_f=1.86\ \mathrm{K\,kg\,mol^{-1}}$.
   3. $T_f^2 = 74529$. …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.A non-volatile solute is dissolved in water. The $\Delta T_b$ of resultant solution is 0.052 K. What is the freezing point of the solution (in K)? ($K_b$ of water = 0.52 K kg mol$^{-1}$; $K_f$ of water = 1.86 K kg mol$^{-1}$; Freezing point of water = 273 K) (A) 272.628 (B) 273.186 (C) 273.000 (D) 272.814

   ›Reveal solutionSolution

   From the boiling-point elevation, the molality is found (0.1 mol/kg), and then the SAME molality is used with Kf to get the freezing-point depression, giving a freezing point of 272.814 K.

   Concept and Intuition

   Both boiling-point elevation and freezing-point depression are colligative properties governed by the same solution molality — ΔTb = Kb·m and ΔTf = Kf·m. Since molality doesn't change, we can find it from one property and use it to compute the other.

   Step-by-Step Solution

   1. ΔTb = Kb·m → m = ΔTb/Kb = 0.052 / 0.52 = 0.1 mol/kg.
   2. ΔTf = Kf·m = 1.86 × 0.1 = 0.186 K.
   3. Freezing point of solution = freezing point of pure water − ΔTf = 273 − 0.186 = 272.814 K. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.The following graph is obtained for vapour pressure (in atm) (on y-axis) and T (in K) (on x-axis) for aqueous urea solution and water. What is the boiling point (in K) of urea solution? (Atmospheric pressure = 1 atm) [FIGURE] (two vapour-pressure-vs-temperature curves rising left to right; three horizontal dashed lines at y = 0.75, 1.00 and 1.25 atm intersect the two curves; the resulting four intersection points are projected onto the x-axis at four temperatures labeled, in increasing order, $T_1, T_2, T_3, T_4$) (A) $T_1$ (B) $T_2$ (C) $T_3$ (D) $T_4$

   ›Reveal solutionSolution

   The solution's boiling point is where its vapour-pressure curve meets $P=1$ atm, which is $T_3$ on the graph (higher than pure water's $T_2$, consistent with boiling point elevation).

   Concept and Intuition

   A liquid boils at the temperature where its vapour pressure equals the surrounding atmospheric pressure. Dissolving a non-volatile solute (urea) lowers the solution's vapour pressure at any given temperature (Raoult's law), which means the solution's vapour-pressure curve sits to the right of pure water's curve — it needs a HIGHER temperature to reach the same vapour pressure. This is exactly boiling point elevation, $\Delta T_b > 0$.

   Step-by-Step Solution

   1. Atmospheric pressure is given as 1 atm, so the boiling point of any liquid on this graph is where its curve crosses the horizontal $y=1.00$ line.
   2. The problem states the left curve (pure water, lower vapour pressure needed at lower temperature) crosses $y=1.00$ at $T_2$.
   3. The right curve (urea solution, shifted to higher temperature for the same vapour pressure due to boiling point elevation) crosses $y=1.00$ at $T_3$. …

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9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.The elevation in the boiling point of aqueous urea solution is 0.104 K. What is its $\Delta T_f$ (in K) value? (for Water $K_b = 0.52$ K kg mol$^{-1}$, $K_f = 1.86$ K kg mol$^{-1}$) (A) 0.0186 (B) 0.186 (C) 0.372 (D) 0.0372

   ›Reveal solutionSolution

   The same molality drives both boiling-point elevation and freezing-point depression; find $m$ from $\Delta T_b$, then use it with $K_f$ to get $\Delta T_f$.

   Concept and Intuition

   Both colligative properties depend on the same solution molality $m$: $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$. Since urea is a non-electrolyte (no dissociation), the molality computed from one property applies directly to the other.

   Step-by-Step Solution

   1. $m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.104}{0.52} = 0.2\ \text{mol/kg}$. …

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10. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.What is the boiling point of solution of 0.1m KCl? $K_b$ of water is 0.52 K kg mol$^{-1}$. ($\alpha = 100\%$) (water boil at 373 K) (A) 100.104 K (B) 373.104 K (C) 273.104 K (D) 373.052 K

    ›Reveal solutionSolution

    Since KCl fully dissociates into 2 ions per formula unit (van't Hoff factor $i=2$), the boiling point elevation is $\Delta T_b = i K_b m = 0.104$ K, giving a boiling point of 373.104 K.

    Concept and Intuition

    Boiling point elevation is a colligative property that depends on the total number of solute particles in solution, not just the number of formula units dissolved. For an electrolyte like KCl that dissociates completely into K$^+$ and Cl$^-$ ($\alpha = 100\%$), each mole of KCl produces 2 moles of particles, so the van't Hoff factor $i = 2$ must be included in the elevation formula.

    Step-by-Step Solution

    1. Formula: $\Delta T_b = i \, K_b \, m$.
    2. Since KCl dissociates completely ($\alpha=100\%$) into 2 ions (K$^+$ + Cl$^-$), $i = 1 + \alpha(n-1) = 1 + 1\times(2-1) = 2$.
    3. Substitute: $\Delta T_b = 2 \times 0.52 \times 0.1 = 0.104$ K. …

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11. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.What is the depression of freezing point, when mole fraction of non-electrolyte solute in aqueous solution is 0.01? ($K_f$ of $\mathrm{H_2O} = 1.86$ K kg $\mathrm{mol^{-1}}$) (A) 1.246 K (B) 1.380 K (C) 1.528 K (D) 1.043 K

    ›Reveal solutionSolution

    Converting the given mole fraction of solute to molality and applying $\Delta T_f=K_f m$ gives a freezing-point depression of about 1.043 K.

    Concept and Intuition

    Freezing point depression depends on molality, not mole fraction directly, so we must first convert. For a solution with total 1 mole (basis), if $x_2=0.01$ is the solute's mole fraction, then $n_2=0.01$ and $n_1=0.99$ (moles of water). Molality is moles of solute per kg of solvent.

    Step-by-Step Solution

    1. Take a basis of 1 total mole: $n_2 = 0.01$ mol solute, $n_1 = 0.99$ mol water.
    2. Mass of water (solvent): $0.99\ \text{mol}\times18\ \text{g/mol} = 17.82$ g $=0.01782$ kg.
    3. Molality $m = \dfrac{n_2}{\text{mass of solvent in kg}} = \dfrac{0.01}{0.01782} \approx 0.5612$ mol/kg. …

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12. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.At T (K) x g of a non-volatile solid (molar mass 78 g $\mathrm{mol^{-1}}$) when added to 0.5 kg water, lowered its freezing point by $1.0^{\circ}C$. What is x (in g)? ($K_f$ of water at T(K) = 1.86 K Kg $\mathrm{mol^{-1}}$) (A) 10.48 (B) 20.96 (C) 41.92 (D) 5.24

    ›Reveal solutionSolution

    Freezing-point depression gives the molality directly; converting molality to mass via the given molar mass gives $x \approx 20.96\,g$.

    Concept and Intuition

    Freezing-point depression is a colligative property: $\Delta T_f = K_f \times m$, where $m$ is the molality of the solute (moles of solute per kg of solvent) and $K_f$ is the cryoscopic constant of the solvent. Since the solute is non-volatile and (implicitly) a non-electrolyte (no van't Hoff factor mentioned), we use this formula directly.

    Step-by-Step Solution

    1. Given: $\Delta T_f = 1.0^{\circ}C$, $K_f = 1.86\ K\,kg\,mol^{-1}$, mass of water $=0.5\ kg$, molar mass of solute $M = 78\ g\,mol^{-1}$.
    2. Find molality: $m = \dfrac{\Delta T_f}{K_f} = \dfrac{1.0}{1.86} = 0.5376\ mol\,kg^{-1}$.
    3. Molality is moles of solute per kg solvent, so moles of solute $n = m \times (\text{mass of water in kg}) = 0.5376 \times 0.5 = 0.2688\ mol$. …

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