Question

$1.46g$ of a biopolymer dissolved in a $100mL$ water at $300K$ exerted an osmotic pressure of $2.42\times {10}^{-3}$ bar.

The molar mass of the biopolymer is $x\times {10}^{4}g{\text{mol}}^{-1}$ (Round off to the Nearest Integer)

• A. 15.00
• B. 15
• C. 15.0

Answer 1

Answer: (A), (B), (C)

Osmotic pressure: $\left(\pi \right)$
$\pi =i\times C\times R\times T$
Here
$i\text{is van't Hoff factor}C\text{is concentration (Molarity)}R\text{is ideal gas constant}T\text{is temperature}$

$2.42\times {10}^{-3}=\frac{1\times 1.46\times 1000\times 0.083\times 300}{M\times 100}$

$M=150223g{\text{mol}}^{-1}$

$M=15.0223\times {10}^4{\text{g mol}}^{-1}x=15$