The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg-1) of the aqueous solution is

Question

The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg-1) of the aqueous solution is (A) 13.88 × 10-1 (B) 13.88 × 10-2 (C) 13.88 (D) 13.88 × 10-3

Tardigrade Admin · Accepted Answer

X textsolvent = 0.8 If nT =1 n textSolvent = 0.8 n textSolute =0.2 molality = (0.2/(0.8×18)1000) = 13.88

Q. The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol $k g^{- 1}$ ) of the aqueous solution is

$13.88 \times 1 0^{- 1}$

$13.88 \times 1 0^{- 2}$

$13.88$

$13.88 \times 1 0^{- 3}$

$X_{solvent} = 0.8$
If $n_{T} = 1$
$n_{Solvent} = 0.8$
$n_{Solute} = 0.2$
molality $= \frac{0.2}{\frac{0.8 \times 18}{1000}} = 13.88$

Q. The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg−1) of the aqueous solution is

13.88×10−1

13.88×10−2

13.88

13.88×10−3