Question

$100\, mL$ of a solution containing $0.4\, g$ urea $\left( NH _{2} CONH _{2}\right)$ and $0.9 \,g$ of another substance $( X )$ shows an osmotic pressure of $2.85$ atm at $25^{\circ} C$. Molar mass of the substance $( X )$ is

• A. $85\, g\, mol ^{-1}$
• B. $92 \, g\, mol ^{-1}$
• C. $180\, g\, mol ^{-1}$
• D. $140\, g\, mol ^{-1}$

Answer 1

Answer: (C)

Osmotic pressure, $\pi=\left(M_{1}+M_{2}\right) R T$

$M_{1}$ ( urea ) $=\frac{0,4 \,g}{60 \,g\, mol ^{-1} \times 0.1 \,L }$

$=\frac{1}{15} \,mol\, L ^{-1}$

$M_{2}(X)=\frac{0.9 \,g }{m \,g\, mol ^{-1} \times 0.1 \,L }$

$=\frac{8}{m} \,mol\, L ^{-1}$

$\pi=2.85$ atm (given)

$\therefore 2.85=\left(\frac{1}{15}+\frac{9}{m}\right) \times 0.0821 \times 298$

$\left(\frac{1}{15}+\frac{9}{m}\right)=0.1165$

$\frac{9}{m}=\left(0.1165-\frac{1}{20}\right)$

$\frac{9}{m}=0.0498$

$\therefore m =180 \,g\, mol ^{-1}$