Question

I need help plz

A hydrocarbon X effuses through a pinhole 1.67 times faster than Kr under the same conditions of temperature and pressure. Calculate the molar mass of the hydrocarbon

Answer 1

Graham's Law :- The rates of effusion of two gases are inversely proportional to the square roots of their molar masses.

Therefore

$\frac{Rate_1}{Rate_2}= \sqrt{\frac{M_2}{M_1}}$

where:

Rate₁ is the rate of effusion of the first gas (volume or number of moles per unit time).

Rate₂ is the rate of effusion for the second gas.

M₁ is the molar mass of gas 1

M₂ is the molar mass of gas 2.

Or

$M_1R_1^2=M_2R_2^2$

Given is:-

Molor mass of Hydrocarbon X is = $M_1$

Molor mass of Kr is = 83.80 g/mol

Rate of effusion of Hydrocarbon X = 1.67

Rate of effusion of Kr = 1

By putting all the values in above equation, we get

$M_1 (1.67)^2=83.8 (1)^2$

$M_1 (2.7889)=83.8$

$M_1 = \frac{83.8}{2.7889}=30.04g/mol$

Therefore the molor mass of Hydrocarbon X is = 30.04g/mol