Question

$300c{m}^3$ of an aqueous solution contains 1 g of a polymer: The osmotic pressure of such solution at 300 K is found to be ${10}^{-3}$ bar.The molar mass of the polymer is?

Answer 1

Do you remember the formula for calculating osmotic pressure?

It is $\pi =C\ast R\ast T$, where $\pi$ is the osmotic pressure.

It can also be written as

$\pi \ast V=n\ast R\ast TasC=\frac{n}{V}$

Here n $=\frac{Givenweight}{Molarweight}$$=\frac{W_b}{M_b}$

Hence, the equation can also be written as

$\pi \ast V=\left(\frac{W_b}{M_b}\right)\ast R\ast T$

In the question, we are given,

the weight of a polymer $W_b$, the temperature T and the osmotic pressure P and the volume V.

We need to find $M_b$.

So, let’s arrange the equation for convenience.

$M_b=\frac{(W_b\ast R\ast T)}{\pi \ast V}$

$W_b=1gm,\pi ={10}^{-3}bar,T=300K,V=300c{m}^3=0.2L,R=0.082$

Plugging in these values we find the molar mass of the polymer to be 82g.