Question

The relative rates of difusion of $C{O}_2$ and $C{l}_2$ gases is $1:1.267$.

• A. True
• B. False

Answer 1

The correct option is A True

We know from Grahams Law of effusion,

$\frac{\text{Rate of}-C{O}_2}{\text{Rate of}-C{l}_2}=\sqrt{\frac{\text{Molar mass of}-C{l}_2}{\text{Molar mass of}-C{O}_2}}$

So,

$\frac{r_{C{O}_2}}{r_{C{l}_2}}=\sqrt{\frac{71}{48}}=1.267$

Hence the given statement is correct