The molecular weight of a gas is 44. the volume occupied by 2.2 g of this gas at `0^(@)C` and 2 atm pressure will be

Options

A. 0.56 lt
B. 1.2 lt
C. 2.4 lt
D. 5.6 lt

Answer

A

Explanation

`PV=nRT`
`V=(nRT)/(P)=(2.2)/(44)xx(273xx8.314)/(2xx10^(5))`
`=(1)/(2)xx(273)/(2)xx10^(-5)xx8.314`
`=(273xx8.314xx10^(-5))/(4)=56.74xx10^(-5)`
`=0.5674xx10^(-3)m^(3)`
=0.5674 lit.