Question

When $31\text{g}$ of non-volatile solute is dissovled in $100\text{g}$ of water, the vapour pressure is lowered by $1.75\text{mm}$. If the vapour pressure of water at $25{}^o\text{C}$ is $17.5\text{mm}$, what is the molecular mass of solute $\left(\text{in g}\right)$?
(Use $\frac{100}{18}=5.55$)

• A. $75\text{g}$
• B. $25\text{g}$
• C. $50\text{g}$
• D. $45\text{g}$

Answer 1

The correct option is C $50\text{g}$

Let molecular mass of the solute is $m_B$.
We know, $p=p_o{x}_A$
$(17.5-1.75)=17.5\times \frac{100/18}{100/18+\frac{25}{m_B}}\Rightarrow \frac{15.75}{17.5}(\frac{100}{18}+\frac{31}{m_B})=\frac{100}{18}\Rightarrow 0.9(5.55+\frac{31}{m_B})=\frac{100}{18}\Rightarrow 5.55+\frac{31}{m_B}=6.17\Rightarrow \frac{31}{m_B}=0.62\Rightarrow m_B=50\text{g}$