Question

$18g$ of glucose is dissolved in $90g$ 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, $w_2=18g,W_1=90g$

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

$\frac{p^{\circ }-p_8}{p^{\circ }}=\frac{18}{180}\times \frac{18}{90}=0.02$