JEE Advanced 2021 Paper 1 · Q09 · Solutions & Colligative Properties

Correct answer:100.10

Solution

For 0.1 molal AgNO$_3$(aq) the dissociation AgNO$_3$ → Ag$^+$ + NO$_3^-$ gives van't Hoff factor i = 2 (1 mol gives 2 ions of solute particles). Boiling-point elevation $\Delta T_b$ = i \* K$_b$ \* m = 2 \* 0.5 \* 0.1 = 0.10 K.

Boiling point of solution A = 100 + 0.10 = 100.10 °C. Therefore x = 100.10.

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