Two beakers, one containing 20 mL of a 0.05 M aqueous solution of a non-volatile, non-electrolyte and the other, the same volume of 0.03 M aqueous solution of NaCl, are placed side by side in a closed enclosure. Volume of the solution (after long time) in the first and second beaker are respectively :

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21.8 mL and 18.2 mL

18.2 mL and 21.8 mL

20 mL and 20 mL

20 mL and 25 mL

answer is B.

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Detailed Solution

$\begin{array}{l} 20 mL, 0.05 M \\ i = 1 \end{array}$ Question Image $\begin{matrix} 20 mL, 0.03 M \\ i = 2 \end{matrix}$

Finally, V.P. will be same in both.

or $\frac{i_{1} n_{1}}{N_{1}} = \frac{i_{2} n_{2}}{N_{2}}$

$\Rightarrow \frac{1 \times 0.05}{(\frac{20 - x}{18})} = \frac{0.03 \times 2}{(\frac{20 + x}{18})} \Rightarrow n = \frac{20}{11}$

Thus, $I \to 20 - \frac{20}{11} = \frac{200}{11} mL$

$II \to 20 + \frac{20}{11} = \frac{240}{11} mL$ .

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