Graham's law of diffusion The rates of diffusion of two gases $\ce{A}$ and $\ce{B}$ are in the ratio $1:4$. If the ratio of their masses present in the mixture is $2:3$, the ratio of their mole fraction is?


My tries:

$\frac{r_1}{r_2}=\frac{1}{4}=\sqrt{\frac{M_2}{M_1}}\to\frac{1}{16}=\frac{M_2}{M_1}$, $M_i$ is molar mass of $i$.

Also given $\frac{w_1}{w_2}=\frac{2}{3}$, multiplying this by above one gives $\frac{M_2\cdot w_1}{M_1\cdot w_2}=\frac{2}{3\cdot 16}\to\frac{n_1}{n_2}=\frac{1}{24}$, where $n_i$ represents number of moles of $i$.

But this is not a correct answer please help.
$$\frac{r_A}{r_B} =\frac{1}{4}\implies\sqrt{\frac{M_B}{M_A}}=\frac{1}{4}$$
$$\therefore \frac{M_B}{M_A} = \frac{1}{16}$$
On dividing the given mass by molar mass, we get the respective number of moles.
Let the number of moles be denoted by $n$
$$\therefore {n_A} = \frac{m_{A}}{M_{A}}$$
$$\therefore {n_B} = \frac{m_{B}}{M_{B}}$$
$$\therefore \frac{n_{A}}{n_B} = \frac{m_{A}\cdot M_{B}}{M_{A}\cdot m_{B}} = \frac{m_{A}}{m_{B}} \cdot \frac{M_{B}}{M_{A}} = \frac{1}{24}$$

I do not see what's wrong.
Do you want the mole fractions of A and B?