Question

AT ${25}^{o}C$, a solution containing 0.2 g of polyisobutylene in 100 mL of benzene developed a rise of 2.4 mm at osmotic equilibrium. The molar mass of polyisobutylene if the density of the solution is 0.88 g/mL is :

• A. $2.39\times {10}^{5}gmo{l}^{-1}$
• B. $7.36\times {10}^{5}gmo{l}^{-1}$
• C. $5.45\times {10}^{5}gmo{l}^{-1}$
• D. None of these

Answer 1

Answer: (A)

$2.39\times {10}^{5}gmo{l}^{-1}$

The osmotic pressure equation is given by

$\pi V=nRTi$

where $n$ is number of moles,

$\pi$ is osmotic pressure,

$i$ is the Vant Hoff factor, which equals $1$ for nonionics

$R=0.08206Latmmo{l}^{-1}{K}^{-1}$

Osmotic pressure in atm:

$\pi =2.4mm\times \left(\frac{0.88mmHg}{13.6mm}\right)\times \left(\frac{1atm}{760mmHg}\right)=2.04\times {10}^{-4}atm$

$\frac{n}{V}=\frac{\pi }{RT}=\frac{2.04\times {10}^{-4}atm}{0.08206Latmmo{l}^{-1}{K}^{-1}\times 298K}=0.83\times {10}^{-5}M$

Amount solute$=100.0mL\times \left(\frac{1L}{1000mL}\right)\times (0.83\times {10}^{-5}M)=0.83\times {10}^{-6}mol$ solute

Molar mass $=\frac{0.2g}{8.3\times {10}^{-6}}mol=240963g/mol=2.4\times {10}^{5}g/mol$