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, \(CO_2\) = 44 g/mole
Molar mass of oxygen gas, \(O_2\) = 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.
\(R\propto \sqrt{\frac{1}{M}}\)
or,
\(\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}\)
where,
\(R_1\) = rate of effusion of a carbon dioxide gas
\(R_2\) = rate of effusion of oxygen gas
\(M_1\) = molar mass of a carbon dioxide gas
\(M_2\) = molar mass of oxygen gas
Now put all the given values in the above formula, we get the molar mass of a gas.
\(\frac{R_1}{R_2}=\sqrt{\frac{31.998}{44}}\)
\(\frac{R_1}{R_2}=0.85\)
\({R_2}=1.17\times {R_1}\)
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.
