The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g $c m^{- 3}$ . The ratio of the molecular weights of the solute and solvent, $(\frac{M W_{s o l u t e}}{M W_{s o l v e n t}})$ , is

Open in App

Solution

$m = \frac{X_{A} \times 1000}{X_{B} \times M_{A}}$

$m = \frac{1000}{9 M_{A}} \dots \dots (i)$

$M = \frac{n_{B} \times 1000 \times d}{n_{A} \times M_{A} + n_{B} \times M_{B}} = \frac{X_{B} \times 1000 \times d}{X_{A} \times M_{A} + X_{B} \times M_{B}}$

$= \frac{200}{0.9 M_{A} + 0.1 M_{B}}$

$= \frac{2000}{9 M_{A} + M_{B}} \dots \dots (i i)$

At m = M

$\frac{1000}{9 M_{A}} = \frac{2000}{9 M_{A} + M_{B}}$

$9 M_{A} + M_{B} = 18 M_{A}$

$∴ 9 M_{A} = M_{B}$

$∴ \frac{M_{B}}{M_{A}} = 9$

Suggest Corrections

Similar questions

(\frac{M W_{s o l u t e}}{M W_{s o l v e n t}})

c m^{3}

0.1

298 K

298 K

2.0 {g cm}^{- 3}

(\frac{m_{solute}}{m_{solvent}})

0.1

298 K

298 K

2.0 g c m^{- 1}

(\frac{M W_{s o l u t e}}{M W_{s o l v e n t}})

Join BYJU'S Learning Program