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

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

Tardigrade Admin · Accepted Answer

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)

= (\frac{100}{18} - n)

Moles of

H_{2} O

in second (concentrated solution)

= (\frac{100}{18} + n)

Mole fractions are equal at equilibrium.

x_{C_{12}} H_{22} O_{11} = \frac{0.01}{( 0.01 + \frac{100}{18} - n )}

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

∴ n = 1.852

∴ χ_{C_{12} H_{2} o_{i l}} = 2.693 \times 1 0^{- 3} \approx 0.002693

Q. A beaker containing 0.010mol of C12​H22​O11​ in 100gH2​O and a beaker containing 0.020 mol of C12​H22​O11​ in 100g H2​O are placed in a chamber and allowed to equilibrate. Thus, mole fraction of C12​H22​O11​ in both solution is

0.00172, 0.00269

0.00269, 0.00172

0.00269, 0.00269

0.00172, 0.00172

Q. A beaker containing $0.010 m o l$ 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