Question

The rates of diffusion of two gases $A$ and $B$ are in ratio $1:4$. If the ratio of their masses present in the mixture is $2.3$. The ratio of their mole fraction is:

Answer 1

Given that

$A$ and $B$ are gases.

So, $M_A$ and $M_B$ are their molar mass.

Which is given that,

$\frac{r_A}{r_B}=\frac{1}{4}\Rightarrow \sqrt{\frac{M_B}{M_A}}=\frac{1}{4}$

$\therefore \frac{M_B}{M_A}=\frac{1}{16}$

$\frac{m_A}{m_B}=\frac{2}{3}$

Now,

On dividing the given mass by molar mass, we get the respective number of moles. Let the number of moles be denoted by $n$

$\therefore n_A=\frac{m_A}{M_A}$

$n_B=\frac{m_A}{M_B}$

$\frac{n_A}{n_B}=\frac{m_A}{m_B}\times \frac{M_B}{M_A}$

$\frac{1\times 2}{3\times 16}=\frac{2}{48}=\frac{1}{24}$

So,

The ratio of mole fraction $1:24$

This is the required solution.