Question

If the density of vapours of a substance of molar mass 18 gm/mole at 1 atm pressure and 500 K is $0.36$ kg m${}^{-3}$, then the value of Z for the vapours is:

[Take R=0.082 L atm]

• A. $\frac{41}{50}$
• B. $\frac{50}{41}$
• C. 1.1
• D. 0.9

Answer 1

The correct option is B $\frac{50}{41}$

$Z=\frac{PV}{nRT}$

$=\frac{PVM}{wRT}$

$=\frac{PM}{\rho RT}$

$Z=\frac{1\times 18}{0.36\times 0.082\times 500}$

$=\frac{50}{41}$

Hence, the correct option is $\text{B}$