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

Molality (m) $= \frac{X_{s o l u t e} \times 1000}{X_{s o l v e n t} \times M_{s o l v e n t}} = \frac{0.1 \times 1000}{0.9 \times M_{s o l v e n t}}$

Molarity (M) $= \frac{0.1 \times 2 \times 1000}{0.1 M_{s o l u t e} + 0.9 M_{s o l v e n t}}$

As Molality $=$ Molarity

So, $\frac{0.1 \times 1000}{0.9 M_{s o l v e n t}} = \frac{0.1 \times 2 \times 1000}{0.1 M_{s o l u t e} + 0.9 M_{s o l v e n t}}$

So, $0.1 M_{s o l u t e} + 0.9 M_{s o l v e n t} = 1.8 M_{s o l v e n t}$

So, $\frac{M_{s o l u t e}}{M_{s o l v e n t}} = 9$

Suggest Corrections

Similar questions

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}})

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

c m^{3}

Join BYJU'S Learning Program