Question

The mass of glucose that should be dissolved in $50g$ of water in order to produce the same lowering of vapour pressure as is produced by dissolving $1g$ 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 $\frac{W_{\text{(Glucose) }}\times 18}{50\times 180}=\frac{1\times 18}{50\times 60}$

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