Exams › JEE Main › Chemistry

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86 °C/m, the freezing point of the solution will be:

1. -0.18 °C
2. -0.54 °C
3. -0.36 °C
4. -0.24 °C

Correct answer: -0.36 °C

Solution

The freezing point depression is given by ΔT = Kf × m, where m is the molality of the solution. Since the solution is 30% ionized, the effective molality is 0.3 × 0.1 = 0.03 m. Therefore, ΔT = 1.86 × 0.03 = 0.0558 °C. Since the solution is below the freezing point of water, the freezing point depression is negative.

Related JEE Main Chemistry questions

• The van't Hoff factor i for a compound which undergoes dissociation in one solvent and association in another solvent is respectively:
• A 100 mL sample of a urea solution contains $6.02 \times 10^{20}$ molecules of urea. What is the molarity of the solution? (Take Avogadro constant, $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$.)
• If the relative atomic mass unit were defined as $\frac{1}{6}$ of the mass of a carbon atom instead of $\frac{1}{12}$, then the mass of one mole of a substance would
• When 4.9 g of $\mathrm{H_2SO_4}$ reacts with NaCl, the products formed are 6 g of sodium hydrogen sulphate and 1.825 g of HCl. What mass of NaCl was decomposed?
• Which of the following has the least mass?
• What amount of magnesium phosphate, $\mathrm{Mg_3(PO_4)_2}$, in moles is needed to provide 0.25 mole of oxygen atoms?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →