Question

$ 18\, g $ of glucose is dissolved in $ 90 \,g $ of water. The relative lowering of vapour pressure of the solution is equal to

• A. 6
• B. 0.2
• C. 5.1
• D. 0.02

Answer 1

Answer: (D)

From Raoult's law for non-volatile solute, we know that

$\frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}$

For dilute solution, it becomes

$\frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}}=\frac{w_{2}}{m_{2}} \times \frac{M_{1}}{W_{1}}$

Given that, $w_{2}=18\, g , W_{1}=90 \,g$

Molecular mass of glucose

$\left( C _{6} H _{12} O _{6}\right)=12 \times 6+12 \times 1+6 \times 16=180$

Molecular mass of water

$\left( H _{2} O \right)=2+16=18$

Now, putting the values, we get

$\frac{p^{\circ}-p_{8}}{p^{\circ}}=\frac{18}{180} \times \frac{18}{90}=0.02$