The molar mass of unknown gas is 63.5 g/mol
calculation
let the unknown gas be represent by letter Y
The rate of effusion of a gas varies inversely with square root of molar mass
that is R(CH4)/ R(Y)= √[(M(Y)/M(CH4)]
= 47.8 ml/min/24.0ml/min= √[(Y/16g/mol)]
square both side to remove square root sign
=47.8²ml/min /24²ml/min= Y/16 g/mol
=2284.84ml/min/576ml/min=Y/16g/mol
multiply both side by 16
y =63.5 g/mol
The molar mass of unknown gas whose rate of effusion is 24ml/min is 63.5 g/mol.
Relation between the molar mass of any gas and rate of effusion is defined as:
Rate of effusion ∝ 1/√M
For the given question, equation for the molar mass is written as:
R₁/R₂ = √M₂/M₁, where
R₁ = rate of effusion of methane gas = 47.8 ml/min (given)
R₂ = rate of effusion of unknown gas = 24.0 ml/min (given)
M₁ = molar mass of methane gas = 16g/mol (known)
M₂ = molar mass of unknown gas = to find?
On putting values on the above equation, we get
47.8 ml/min/24.0ml/min= √[(M₂/16g/mol)]
On squaring both sides we get,
47.8²ml/min /24²ml/min= M₂/16 g/mol
2284.84ml/min/576ml/min=M₂/16g/mo
M₂ = 63.5 g/mol
Hence, 63.5 g/mol is the molar mass.
To know more about rate of effusion, visit the below link:
https://brainly.com/question/13547979
