Question

A solution containing $10gd{m}^{-3}$ of urea (Molecular weight 60 g/mol) is isotonic with a $5\%\left(w/V\right)$ solution of a non - volatile solute at the same temperature. The molecular mass of this non volatile solute is :

• A. $30gmo{l}^{-1}$
• B. $50gmo{l}^{-1}$
• C. $111.1gmo{l}^{-1}$
• D. $300gmo{l}^{-1}$

Answer 1

The correct option is D $300gmo{l}^{-1}$

$5\%\left(w/V\right)$ solution of a non - volatile solute means:
$5g$ of non volatile solute is present in $100mL$ of solution
For isotonic solutions:

${\pi }_{\left(\text{urea}\right)}={\pi }_{\text{(unknown solute)}}{C}_{\left(\text{urea}\right)}\times R\times T=C_{\text{(unknown solute)}}\times R\times T$

$C_{\left(\text{urea}\right)}=C_{\left(\text{unknown solute}\right)}$

$\because 1d{m}^3=1L$
So,
$C_{urea}=\frac{10}{60}M$
and
$C_{\text{unknown solute}}=\frac{5\times 1000}{m_B\times 100}$

So putting the values,

$\frac{10}{60}=\frac{5\times 1000}{m_B\times 100}$

$m_B=300gmo{l}^{-1}$