JEE Main 2026 — Chemistry Physical Chemistry

The molality of the solution is given by:

$$m = \dfrac{\text{Mass of solute}}{\text{Molar mass of solute} \times \text{Mass of solvent in kg}}$$m=Molar mass of solute×Mass of solvent in kgMass of solute​

$$m = \dfrac{19.5}{78 \times 0.5} = 0.5 \text{ mol kg}^{-1}$$m=78×0.519.5​=0.5 mol kg−1

The depression in freezing point is given by:

$$\Delta T_f = i \times K_f \times m$$ΔTf​=i×Kf​×m

Substituting the given values:

$$1 = i \times 1.86 \times 0.5$$1=i×1.86×0.5

$$1 = i \times 0.93$$1=i×0.93

$$i = \dfrac{1}{0.93} = \dfrac{100}{93}$$i=0.931​=93100​

For the dissociation of a weak acid $HA \rightleftharpoons H^+ + A^-$HA⇌H++A−, the van 't Hoff factor is $i = 1 + \alpha$i=1+α, where $\alpha$α is the degree of dissociation.

$$1 + \alpha = \dfrac{100}{93}$$1+α=93100​

$$\alpha = \dfrac{100}{93} - 1 = \dfrac{7}{93}$$α=93100​−1=937​

The acid dissociation constant $K_a$Ka​ is given by:

$$K_a = \dfrac{C \alpha^2}{1 - \alpha}$$Ka​=1−αCα2​

Since molarity and molality have the same values, $C = 0.5 \text{ M}$C=0.5 M.

$$K_a = \dfrac{0.5 \times \left(\dfrac{7}{93}\right)^2}{1 - \dfrac{7}{93}}$$Ka​=1−937​0.5×(937​)2​

$$K_a = \dfrac{0.5 \times \dfrac{49}{93^2}}{\dfrac{86}{93}} = \dfrac{0.5 \times 49}{93 \times 86}$$Ka​=9386​0.5×93249​​=93×860.5×49​

$$K_a = \dfrac{24.5}{7998} \approx 3.06 \times 10^{-3}$$Ka​=799824.5​≈3.06×10−3

The closest value is $3 \times 10^{-3}$3×10−3.

Answer: $3 \times 10^{-3}$3×10−3