Question

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 enclose. What are the volumes in the two beakers when equation is attached ?Volume of the solution in the first and second beaker are respectively.

• A. 21.8 ml and 18.2 mL
• B. 18.2 mL and 21.8 mL
• C. 20 mL and 20 mL
• D. 17.1 mL and 22.9 mL

Answer 1

Answer: (B)

18.2 mL and 21.8 mL

Answer 2

Mole of solute in first beaker = $\frac{0.05 \times 20}{1000}$ = 0.001

mole of solute (Na+ & Cl–) in other beaker =$\frac{2 \times 0.03 \times 20}{1000}$ = 0.0012

conc. of IInd beaker is higher then Ist beaker so water flowes from Ist beaker to IInd beaker till both beaker achieved equal conc. let v volume of water flows from Ist to IInd beaker

so $\frac{0.001}{20 - v} = \frac{0.0012}{20 + v}$

v = 1.8 ml

volume of Ist beaker = 20 – 1.8 = 18.2 ml

volume of IInd beaker = 20 + 1.8 = 21.8 ml.