Question

What will be the molecular mass of $1$ mole of chloroform vapour at STP ($22.4L$), if $0.10g$ of chloroform vapour volume is $18.96$ $c{m}^3$ at STP?

• A. $118.1g$
• B. $0.1g$
• C. $121.1g$
• D. $18.96g$

Answer 1

The correct option is A $118.1g$

At S.T.P, 1 mole of any gas occupies a volume of 22.4 L or 22400 $c{m}^3$.
Volume of vapour is $18.96c{m}^3$
The number of moles of vapour $=\frac{18.96c{m}^3}{22400c{m}^3}=0.00085$ mol.
The mass of vapour is 0.10 g.

The molecular mass of 1 mole vapour will be

$\frac{0.10g}{0.00085mol}=118.1g$