Question

Two samples of gases ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ are at the same temperature. The molecules of ${}^{\prime }{a}^{\prime }$ are travelling $4$ times faster than molecules of ${}^{\prime }{b}^{\prime }$. The ratio of $M_a/M_b$ of their mass will be

• A. $1/4$
• B. $16/1$
• C. $4/1$
• D. $1/16$

Answer 1

Answer: (D)

$1/16$

Answer: D. 1/16

Given that molecules of 'a' are travelling 4 times faster than molecules of 'b'. That means rate of effusion of gas 'a' is 4 times than that of gas 'b'.

That is,

$\frac{r_a}{r_b}=\frac{4}{1}$ where, $r_a$ - rate of effusion of gas 'a' and $r_b$ - rate of effusion of gas 'b'.

According to Graham's Law of Effusion, for two gases 'a' and 'b' with molar masses $M_a$ and $M_b$ respectively ,

$\frac{r_a}{r_b}=\sqrt{\frac{M_b}{M_a}}$

Therefore, we can write as,

$\frac{r_a}{r_b}=\frac{4}{1}=\sqrt{\frac{M_b}{M_a}}$

$=>\sqrt{\frac{M_b}{M_a}}=\frac{4}{1}$

$=>\frac{M_b}{M_a}=\frac{16}{1}$

$=>\frac{M_a}{M_b}=\frac{1}{16}$

Therefore, option D. 1 / 16 is the correct answer.