ExamBro
ExamBro
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

  1. A \( 0.111 \)
  2. B \( 0.021 \)
  3. C \( 0.004 \)
  4. D \( 0.222 \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( 0.111 \)

Step-by-step Solution

Detailed explanation

(A)
\[
\begin{array}{l}
\frac{P^{\circ}-P}{P^{\circ}}=\frac{W_{2}}{M_{2}} \times \frac{M_{1}}{W_{1}} \\
0.002=\frac{W_{2}}{M_{2}} \times \frac{18}{1000} \\
\frac{W_{2}}{M_{2}}=0.11 \text { mole }
\end{array}
\]
\[
\text { Molality }=\frac{W_{2}}{M_{2}} \times \frac{1000}{W_{1}}=0.11 \times \frac{1000}{1000}=0.11 \mathrm{~m}
\]
From KCET
Explore more questions on app