Question

A beaker containing $0.010 \,mol$ of $C _{12} H _{22} O _{11}$ in $100\, g\, H _{2} O$ and a beaker containing $0.020$ mol of $C _{12} H _{22} O _{11}$ in $100 \,g$ $H _{2} O$ are placed in a chamber and allowed to equilibrate. Thus, mole fraction of $C _{12} H _{22} O _{11}$ in both solution is

• A. 0.00172, 0.00269
• B. 0.00269, 0.00172
• C. 0.00269, 0.00269
• D. 0.00172, 0.00172

Answer 1

Answer: (C)

Water vapour will be transferred from the more dilute solution to the more concentrated solution until both solution reach at the same concentration.

Moles of $H _{2} O$ transferred $=n$

Moles of $H _{2} O$ in first (dilute solution)

$=\left(\frac{100}{18}-n\right)$

Moles of $H _{2} O$ in second (concentrated solution)

$=\left(\frac{100}{18}+n\right)$

Mole fractions are equal at equilibrium.

$x_{ C _{12}} H _{22} O _{11}=\frac{0.01}{\left(0.01+\frac{100}{18}-n\right)}$

$=\frac{0.02}{\left(0.02+\frac{100}{18}+n\right)}$

$\therefore n=1.852$

$\therefore \chi_{ C _{12} H _{2} o _{ il }}=2.693 \times 10^{-3} \approx 0.002693$