If ‘a’ gram of non-volatile solute (A) is dissolved in ‘b’ gram of pure liquid (B), the vapour pressure becomes 910th of it’s original value. The molecular weight of the solvent (B) is 15th of the molecular weight of solute (A). Find (b/a)?

P∘−P℘P∘=nN=nAnB=P℘=910P∘P∘−910P∘P∘=WAWB×WBWA110=ab×12⇒ba=5

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If ‘a’ gram of non-volatile solute (A) is dissolved in ‘b’ gram of pure liquid (B), the vapour pressure becomes $\frac{9}{10}$ th of it’s original value. The molecular weight of the solvent (B) is $\frac{1}{5}$ th of the molecular weight of solute (A). Find (b/a)?

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answer is 5.

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Detailed Solution

$\begin{array}{l} \frac{P^{\circ} - P_{℘}}{P^{\circ}} = \frac{n}{N} = \frac{n_{A}}{n_{B}} = P_{℘} = \frac{9}{10} P^{\circ} \\ \frac{P^{\circ} - \frac{9}{10} P^{\circ}}{P^{\circ}} = \frac{W_{A}}{W_{B}} \times \frac{W_{B}}{W_{A}} \\ \frac{1}{10} = \frac{a}{b} \times \frac{1}{2} \Rightarrow \frac{b}{a} = 5 \end{array}$