Question

The mass of glucose that should be dissolved in $50\, g$ of water in order to produce the same lowering of vapour pressure as is produced by dissolving $1\, g$ of urea in the same quantity of water is

• A. 1 g
• B. 3 g
• C. 6 g
• D. 18 g

Answer 1

Answer: (B)

$\frac{P-P_{s}}{P}=\frac{w_{1} M_{2}}{w_{2} M_{1}}$

To produce same lowering of vapour pressure, $\frac{P-P_{s}}{P}$ will be same for both cases.

So, $\frac{W_{\text {(Glucose) }} \times 18}{50 \times 180} =\frac{W_{\text {(urea) }} \times 18}{50 \times 60} $

$W_{\text {(Glucose) }} $ = weight of glucose

$W_{\text {(Glucose) }}$ = weight of urea

or $\frac{W_{\text {(Glucose) }} \times 18}{50 \times 180}=\frac{1 \times 18}{50 \times 60}$

$W_{\text {(Glucose) }}=3$