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 (. Take R =0.082 L atm mole K -1 )

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 (. Take R =0.082 L atm mole K -1 ) (A) (41/50) (B) (50/41) (C) 1.8 (D) 0.9

Tardigrade Admin · Accepted Answer

Vr e a l=(m o l a r m a s s/d e n s i t y)=(18/0.36) Vi d e a l=(n R T/P)=(1 × 0.082 × 500/1) So, Z=(Vr e a l/Vi d e a l)=(50/0.082 × 500)=(50/41)

Q. If density of vapours of a substance of molar mass $18 g / m o l$ at $1 a t m$ pressure and $500 K$ is $0.36 k g m^{- 3}$ , then value of $Z$ for the vapours is
$($ Take $R = 0.082 L$ atm mole $K^{- 1}$ )

$\frac{41}{50}$

$\frac{50}{41}$

$1.8$

$0.9$

$V_{re a l} = \frac{m o l a r ma ss}{d e n s i t y} = \frac{18}{0.36}$
$V_{i d e a l} = \frac{n RT}{P} = \frac{1 \times 0.082 \times 500}{1}$
So, $Z = \frac{V _{re a l}}{V _{i d e a l}} = \frac{50}{0.082 \times 500} = \frac{50}{41}$

Q. 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 ( Take R=0.082L atm mole K−1 )

5041​

4150​

1.8

0.9