Question

A compound exists in the gaseous state both as a monomer (A) and dimer $\left(A_2\right)$. The molecular weight of the monomer is 48. In an experiment, 96 g of the compound was confined in a vessel of volume 33.6 litres and heated to ${273}^{\circ }C$. Calculate the pressure developed, if the compound exists as a dimer to the extent of 50 per cent by weight, under these conditions.

[Use R= 0.082 atm -litre/ mole .K].

Answer 1

Since A and $A_2$ are two states in the gaseous phase having their weight ratio $50$ i.e., $1:1$

$\therefore$ Mole of A$=\frac{96}{2}\times \frac{1}{48}=1$
We have $n=\frac{w}{m}$

$\therefore$Mole of $A_2=\frac{96}{2}\times \frac{1}{96}=\frac{1}{2}$

$\therefore$ Total mole of A and $A_2$ are$=1+\frac{1}{2}=\frac{3}{2}$

We know that $PV=nRT$

$P\times 33.6=\frac{3}{2}\times 0.0821\times 546$

$P=2$ atm