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JEE Advance - Chemistry (2021 - Paper 1 Online - No. 10)

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 | y | is ________.
Answer
2.5

Explanation

$$ \begin{aligned} & \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{NO}_3^{-}(\mathrm{aq}) \\\\ & 0.05 \mathrm{~m} \quad\quad\quad\quad\quad 0.05 \mathrm{~m} \quad\quad 0.05 \mathrm{~m} \\\\ & \mathrm{BaCl}_2(\mathrm{aq}) \longrightarrow \mathrm{Ba}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \\\\ & 0.05 \mathrm{~m} \quad \quad\quad\quad\quad0.05 \mathrm{~m} \quad\quad 0.1 \mathrm{~m} \end{aligned} $$

$\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$combine to form $\mathrm{AgCl}$ precipitate

$$ \begin{array}{cccc} & \mathrm{Ag}^{+}(\mathrm{aq})+ & \mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow & \mathrm{AgCl}(\mathrm{s}) \\\\ \mathrm{t}=0 & 0.05 \mathrm{~m} & 0.1 \mathrm{~m} \\\\ \mathrm{t}=\infty & 0 & 0.05 \mathrm{~m} \end{array} $$

In final solution total concentration of all ions :

$$ \begin{aligned} & {\left[\mathrm{Cl}^{-}\right]+\left[\mathrm{NO}_3^{-}\right]+\left[\mathrm{Ba}^{2+}\right]=0.05+0.05+0.05} \\\\ & \begin{aligned} \Delta \mathrm{T}_{\mathrm{b}} & =0.5 \times 0.15 \\\\ & =0.15 \mathrm{~m} \\\\ & =0.075^{\circ} \mathrm{C} \end{aligned} \end{aligned} $$

B.P. of solution ' $\mathrm{B}$ ' $=100.075^{\circ} \mathrm{C}$

B.P. of solution ' $\mathrm{A}^{\prime}=100.1^{\circ} \mathrm{C}$

$|y|=100.1-100.075$

$$ =0.025=2.5 \times 10^{-2} $$

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