Question

The osmotic pressure of a solution of an organic substance containing $18g$ in one litre of solution at $300K$ is $2.7\times {10}^{5}Pa$ .Find the molecular mass of the substance.
Take $R=8.3Jmo{l}^{-1}{K}^{-1}$

• A. $166gmo{l}^{-1}$
• B. $65gmo{l}^{-1}$
• C. $212gmo{l}^{-1}$
• D. $189gmo{l}^{-1}$

Answer 1

The correct option is A $166gmo{l}^{-1}$

Osmotic pressure:

$\pi =\frac{n}{V}\times RT$

Here n is number of moles

$n=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{w_B}{m_B}$

$V=1L={10}^{-3}{m}^{3}R=8.3Jmo{l}^{-1}{K}^{-1}T=300K\pi =2.7\times {10}^{5}Pa{w}_B=18g$

$\Rightarrow \pi V=\frac{w_B}{m_B}\times RT{m}_B=\frac{w_B\times R\times T}{\pi \times V}$

$m_B=\frac{18\times 8.3\times 300}{2.7\times {10}^5\times {10}^{-3}}{m}_B=166gmo{l}^{-1}$