Question

The rate of diffusion of 2 gases A and B are in the ratio 16:3. if the ratio of their messes asert m tre
mixture is 2:3. Then
(A) The ratio of their molar masses is 16:1
B) The ratio of their molar masses is 1.4
(C) The ratio of their moles present inside the container is 1:24
D) The ratio of their moles present inside the container is 3:3
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Answer 1

answer : option (C) The ratio of their moles present inside the container is 1:24

you did mistake to write ratio of their rate of diffusion , it is 1 : 4 not 16 : 3.

using formula,

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

where r denotes rate of diffusion of gases and M denotes molar mass of gases.

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

so, 1/4 = $\sqrt{\frac{M_B}{M_A}}$

or, 1/16 = $\frac{M_B}{M_A}$.....(1)

given, ratio of their masses in the mixture is 2 : 3

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

or, $\frac{\frac{m_A}{M_A}M_A}{\frac{m_B}{M_B}M_B}=\frac{2}{3}$

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

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

from equation (1),

or, $\frac{n_A}{n_B}=\frac{2}{3}\times\frac{1}{16}=\frac{1}{24}$

hence, ratio of their moles present inside the container is 1 : 24.

Answer 2

Answer:

answer is B

Explanation:

RA/RB = 16/2

WA/WB = 2/3

RA/RB=WA/WB * MB/MA* root MB/MA

16/3 = 2/3 * [MB/MA]^3/2

8 =[MA/MB]^3/2

MB/MA = 4/1= 4:1