KCET · Chemistry · Solutions
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
- A \( 0.111 \)
- B \( 0.021 \)
- C \( 0.004 \)
- D \( 0.222 \)
Answer & Solution
Correct Answer
(A) \( 0.111 \)
Step-by-step Solution
Detailed explanation
(A)
\[
\begin{array}{l}
\frac{P^{\circ}-P}{P^{\circ}}=\frac{W_{2}}{M_{2}} \times \frac{M_{1}}{W_{1}} \\
0.002=\frac{W_{2}}{M_{2}} \times \frac{18}{1000} \\
\frac{W_{2}}{M_{2}}=0.11 \text { mole }
\end{array}
\]
\[
\text { Molality }=\frac{W_{2}}{M_{2}} \times \frac{1000}{W_{1}}=0.11 \times \frac{1000}{1000}=0.11 \mathrm{~m}
\]
\[
\begin{array}{l}
\frac{P^{\circ}-P}{P^{\circ}}=\frac{W_{2}}{M_{2}} \times \frac{M_{1}}{W_{1}} \\
0.002=\frac{W_{2}}{M_{2}} \times \frac{18}{1000} \\
\frac{W_{2}}{M_{2}}=0.11 \text { mole }
\end{array}
\]
\[
\text { Molality }=\frac{W_{2}}{M_{2}} \times \frac{1000}{W_{1}}=0.11 \times \frac{1000}{1000}=0.11 \mathrm{~m}
\]
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