QJEEAdv21Ch88d4: JEE Advanced 2021 Chemistry - Class 12 JEEAdv21Ch88d4 Solution: The value of

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

$\mathrm{Ag}^{+}$ 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 ' $\mathrm{A}^{\prime}=100.1^{\circ} \mathrm{C}$

$|y|=100.1-100.075$

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