Question

If density of vapours of a substance of molar mass $18g/mol$ at $1atm$ pressure and $500K$ is $0.36kgm^{- 3}$ , then value of $Z$ for the vapours is
$\left(\right.$ Take $R =0.082 \quad L$ atm mole $K ^{-1}$ )

• A. $\frac{41}{50}$
• B. $\frac{50}{41}$
• C. $1.8$
• D. $0.9$

Answer 1

Answer: (B)

$V_{r e a l}=\frac{m o l a r \, m a s s}{d e n s i t y}=\frac{18}{0.36}$
$V_{i d e a l}=\frac{n R T}{P}=\frac{1 \times 0.082 \times 500}{1}$
So, $Z=\frac{V_{r e a l}}{V_{i d e a l}}=\frac{50}{0.082 \times 500}=\frac{50}{41}$