Question

The mass of 2.24×10-³ m³ of a gas is 4.4 g at 273.15K and 101.kPa pressure.the gas may be

Answer 1

Answer: The gas may be carbon dioxide.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the gas = 101 kPa = 0.99 atm

V= Volume of the gas = $2.24\times 10^{-3}m^3=2.24L$   $1m^3=1000L$

T= Temperature of the gas = 273.15 K    

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= ?

$n=\frac{PV}{RT}=\frac{0.99\times 2.24}{0.0821\times 273.15}=0.1moles$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Molar mass}=\frac{\text{Given mass}}{\text {Number of moles}}=\frac{4.4}{0.1}=44g/mol$

Thus the gas may be $CO_2$ with molecular formula of 44 g/mol.