JEE Advanced 2021 Paper 1 · Q10 · Solutions & Colligative Properties

Question

The value of |y| is ___.

SolveKar AI · Accepted Answer

Mix 1 L (~1000 g) of solution A (0.1 molal AgNO$_3$) with 1 L (~1000 g) of 0.1 molal BaCl$_2$. Initial moles of AgNO$_3$ = 0.1, BaCl$_2$ = 0.1.

Reaction: BaCl$_2$ + 2 AgNO$_3$ → 2 AgCl(s) + Ba(NO$_3$)$_2$. AgNO$_3$ is limiting (needs 0.05 mol BaCl$_2$ to consume 0.1 mol AgNO$_3$), so all 0.1 mol AgNO$_3$ reacts.

After reaction (in 2000 g of solvent, since densities equal water): AgCl precipitates (0.1 mol); remaining BaCl$_2$ = 0.05 mol; Ba(NO$_3$)$_2$ formed = 0.05 mol.

Molality contribution from remaining solutes: BaCl$_2$ → Ba$^{2+}$ + 2 Cl$^-$ (i = 3): 3 \* 0.05/2 kg = 0.075 mol kg$^{-1}$. Ba(NO$_3$)$_2$ → Ba$^{2+}$ + 2 NO$_3^-$ (i = 3): 3 \* 0.05/2 kg = 0.075 mol kg$^{-1}$. Total effective molality = 0.15 mol kg$^{-1}$.

$\Delta T_b$ for B = K$_b$ \* m$_{eff}$ = 0.5 \* 0.15 = 0.075 K. So boiling point of B = 100.075 °C.

Difference: $\Delta T$ = T$_{b,A}$ - T$_{b,B}$ = 100.10 - 100.075 = 0.025 = 2.5 \* 10$^{-2}$ °C. Therefore |y| = 2.5.