Question

The solution containing $4.0g$ of a polyvinyl chloride polymer in 1 liter of dioxane was found to have an osmotic pressure $6.0 \times {10^{ - 4}}atmosphere$ at $300k$, the value of $R$ is $0.082{\text{litre atmosphere mol}}{{\text{e}}^{ - 1}}{k^{ - 1}}$ . The molecular mass of the polymer is found to be
A.$3.0 \times {10^2}$
B.$1.6 \times {10^5}$
C.$5.6 \times {10^4}$
D.$6.4 \times {10^2}$

Answer 1

Hint: We can find the molecular mass of the polymer by using the given value of osmotic pressure. This is the pressure needed to be added to the solution to literally avoid the leakage of the liquid from the liquid to the solution side through the semipermeable membrane. Osmotic strain is inversely proportional to the gram molecular mass.
Formula used: The formula used to find this question’s answer is formula for osmotic pressure. That is,
$\Pi V = nRT$ where,
 $\Pi = $ osmotic pressure
$V = $ volume of solvent
$n = $ number of moles
$R = $gas constant
$T = $Temperature

Complete step by step answer:
Let us analyze the given data;
Given mass of poly vinyl chloride $(W) = 4.0g$
$\Pi = 6.0 \times {10^{ - 4}}atm$
$T = 300k$
$R = 0.082{\text{ L atm mo}}{{\text{l}}^{ - 1}}{k^{ - 1}}$
We have to find the molecular mass $(M)$, From the formula $\Pi \times V = n \times R \times T$
$n = \dfrac{W}{M}$ where $W$is the given mass and $M$is the molecular mass. So, the formula becomes;
$\Pi V = \dfrac{W}{M} \times R \times T$
From this $M = \dfrac{{W \times R \times T}}{{\Pi \times V}}$ so by substituting the given values
$M = \dfrac{{4.0g \times 0.082{\text{ L atm mo}}{{\text{l}}^{ - 1}}{k^{ - 1}} \times 300k}}{{6.0 \times {{10}^{ - 4}}atm \times 1{\text{ }}liter}}$
$ = 1.6 \times {10^5}g/mol$
So, the right option is B.


Additional information:
Osmotic pressure is the minimum pressure needed to be applied to a solvent to prevent its flow through a semipermeable membrane. Osmotic pressure is a colligative property so it will depend on the number of particles present. The term 'osmosis' refers to the passage of solvent molecules from a region where the concentration of the solute is low to a region where the concentration of the solute is high across a semipermeable membrane. The semi-permeable membrane only requires liquid molecules to travel through it; it is difficult for solute particles to migrate through it.

Note:
Since osmotic pressure is a colligative property the van’t hoff factor(i) also needs to be used whenever necessary. Where $i = \dfrac{{{\text{actual value of concentration of particle}}}}{{{\text{observed value of concentration of particle}}}}$.
So, the overall formula become $\Pi V = niRT$