Answer : The correct option is, (C) 1.173
Solution : Given,
Molar mass of carbon = 12.011 g/mole
Molar mass of oxygen = 15.999 g/mole
Molar mass of carbon dioxide gas, [tex]CO_2[/tex] = 44 g/mole
Molar mass of oxygen gas, [tex]O_2[/tex] = 31.998 g/mole
According to the Graham's law, the rate of effusion of a gas is inversely proportional to the square root of the molar mass of a gas.
[tex]R\propto \sqrt{\frac{1}{M}}[/tex]
or,
[tex]\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}[/tex]
where,
[tex]R_1[/tex] = rate of effusion of a carbon dioxide gas
[tex]R_2[/tex] = rate of effusion of oxygen gas
[tex]M_1[/tex] = molar mass of a carbon dioxide gas
[tex]M_2[/tex] = molar mass of oxygen gas
Now put all the given values in the above formula, we get the molar mass of a gas.
[tex]\frac{R_1}{R_2}=\sqrt{\frac{31.998}{44}}[/tex]
[tex]\frac{R_1}{R_2}=0.85[/tex]
[tex]{R_2}=1.17\times {R_1}[/tex]
Therefore, the 1.17 times greater is the rate of effusion of oxygen gas than that of carbon dioxide gas at the same temperature and pressure.
