The mole fraction of a solute in solution is 0.1. A t 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 weight of the solute and solvent, $(\frac{M W_{s o l u t e}}{M W_{s o l v e n t}})$ , is

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answer is 9.

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Detailed Solution

$\begin{array}{l} X_{B} = 0.1 = \frac{n_{B}}{n_{A} + n_{B}} \\ n_{B} = 0.1, n_{A} = 0.9 \\ m = M \Rightarrow 1000 ml \Rightarrow 2000 g \\ \Rightarrow 1000 g of solute + 1000 g of solvent \\ X_{B} = 0.1 = \frac{\frac{1000}{M_{B}}}{\frac{1000}{M_{A}} + \frac{1000}{M_{B}}} \\ \Rightarrow \frac{100}{M_{A}} + \frac{100}{M_{B}} = \frac{1000}{M_{B}} \\ \Rightarrow \frac{1}{M_{A}} + \frac{1}{M_{B}} = \frac{10}{M_{B}} \\ \Rightarrow \frac{1}{M_{A}} = \frac{9}{M_{B}} \\ \Rightarrow \frac{M_{B}}{M_{A}} = 9 \end{array}$