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JEE Advance - Chemistry (2021 - Paper 1 Online - No. 9)

The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x$$^\circ$$C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y $$\times$$ 10$$-$$2$$^\circ$$C.

(Assume : Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.

Use : Molal elevation constant (Ebullioscopic Constant), Kb = 0.5 K kg mol$$-$$1; Boiling point of pure water as 100$$^\circ$$C.)
The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x$$^\circ$$C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y $$\times$$ 10$$-$$2$$^\circ$$C.

(Assume : Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.

Use : Molal elevation constant (Ebullioscopic Constant), Kb = 0.5 K kg mol$$-$$1; Boiling point of pure water as 100$$^\circ$$C.)
The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x$$^\circ$$C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y $$\times$$ 10$$-$$2$$^\circ$$C.

(Assume : Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.

Use : Molal elevation constant (Ebullioscopic Constant), Kb = 0.5 K kg mol$$-$$1; Boiling point of pure water as 100$$^\circ$$C.)
The value of x is ________.
Answer
100.1

Explanation

$$ \begin{aligned} & \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \underset{0.1 ~ m}{\mathrm{Ag}^{+}(\mathrm{aq})}+\underset{0.1 ~m}{\mathrm{NO}_3^{-}(\mathrm{aq})}\\\\ & \Delta \mathrm{T}_{\mathrm{b}}=0.2 \times 0.5 \\\\ & =0.1^{\circ} \mathrm{C}=0.1 \mathrm{~K} \end{aligned} $$

Boiling point of solution $=100.1^{\circ} \mathrm{C} =\mathrm{X} $

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