Question

$1.46\, g$ of a biopolymer dissolved in a $100\, mL$ water at $300\, K$ exerted an osmotic pressure of $2.42 \times 10^{-3}$ bar The molar mass of the biopolymer is________ $-10^{4} gmol ^{-1}$ (Round off to the Nearest Integer)
$\left[\right.$ Use : $R=0.083\, L$ bar $\left.mol^{-1} K^{-1}\right]$

Answer 1

$\pi=C R T$
$\pi=$ osmotic pressure
$C=$ molarity
$T=$ Temperature of solution
let the molar mass be $M gm / mol$
$2.42 \times 10^{-3}$ bar
$=\frac{\left(\frac{1.46 g}{M\, g\, m / mol}\right)}{0.1 l} \times\left(\frac{0.083\, l-bar}{mol-K}\right) \times(300\, K )$
$\Rightarrow M=15.02 \times 10^{4} g / mol$