Q11 / 180 · KCET 2019 · Chemistry
Relative lowering of vapour pressure of a dilute solution of glucose dissolved in \( 1 \mathrm{~kg} \) of water is
\( 0.002 \). The molality of the solution is
\( 0.002 \). The molality of the solution is
- Correct answerA\( 0.111 \)
- B\( 0.021 \)
- C\( 0.004 \)
- D\( 0.222 \)
Solution
(A)
\[ \begin{array}{l} \frac{P^{\circ}-P}{P^{\circ}}=\frac{W_{2}}{M_{2}} \times \frac{M_{1}}{W_{1}} \\ 0.002=\frac{W_{2}}{M_{2}} \times \frac{18}{1000} \\ \frac{W_{2}}{M_{2}}=0.11 \text { mole } \end{array} \]
\[ \text { Molality }=\frac{W_{2}}{M_{2}} \times \frac{1000}{W_{1}}=0.11 \times \frac{1000}{1000}=0.11 \mathrm{~m} \]
\[ \begin{array}{l} \frac{P^{\circ}-P}{P^{\circ}}=\frac{W_{2}}{M_{2}} \times \frac{M_{1}}{W_{1}} \\ 0.002=\frac{W_{2}}{M_{2}} \times \frac{18}{1000} \\ \frac{W_{2}}{M_{2}}=0.11 \text { mole } \end{array} \]
\[ \text { Molality }=\frac{W_{2}}{M_{2}} \times \frac{1000}{W_{1}}=0.11 \times \frac{1000}{1000}=0.11 \mathrm{~m} \]
Hence, the correct answer is option: 'A'.