Question

The rates of diffusion of two gases A and 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: (9^1/3= 2.08)​

Answer 1

Answer:

The ratio of their mole fraction is 1 : 24

Explanation:

The rate of diffusion of gases A and B depends upon their molar mass as

$\boxed{\frac{r_A}{r_B} =\sqrt{\frac{M_B}{M_A}}}$

where $r_A$ and $r_B$ are rates of diffusion of gases A and B and $M_A$ , $M_B$ are their molar masses (i.e. molecular weights)

Given

${\frac{r_A}{r_B} =\frac{1}{4}$

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

If the masses of gases A and B are $m_A$ and $m_B$ then

No. of moles of A

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

Similarly, no. of moles of B

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

Therefore,

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

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

$\implies \frac{n_A}{n_B}=\frac{2}{3}\times \frac{M_B}{M_A}$

$\implies \frac{M_B}{M_A}=\frac{3}{2}\frac{n_A}{n_B}$

Also,

$\frac{M_B}{M_A} =(\frac{r_A}{r_B})^2$

or, $\frac{3}{2}\frac{n_A}{n_B}=(\frac{1}{4})^2$

or, $\frac{n_A}{n_B}=\frac{1}{16}\times \frac{2}{3}$

$\implies \frac{n_A}{n_B}=\frac{1}{24}$

Hope this helps.

Answer 2

Answer:

Here:

M=molarmass

m=mass present

n=number of moles

r= rate