Question

20.
AIIMS 2018
Initially in a container 1 g of gas A has 4 atm
pressure at constant temperature. If 2 g of gas B is
added in same container at same temperature then
pressure becomes 6 atm what will be the ratio of
molecular weight of A and B :-
(1) M = 4M
(2) MA = 2M,
(3) M, = 2M, (4) M, = 4M​

Answer 1

Answer:

The ratio of  molecular weight of A and B is 1:4.

Explanation:

Initially in container 1 g of gas A was present with total pressure of 4 atm

Mass of gas A = 4 g

Molecular weight of gas A = $M_a$

Moles of gas A = $n_a=\frac{1 g}{M_a}$

Partial pressure = Total pressure of gas A = p  = 4 atm

Finally of addition of 2 g of gas B was added in same container and total pressure changes to 6 atm from 4 atm.

Moles of gas B = $n_b=\frac{1 g}{M_b}$

Molecular weight of gas B = $M_b$

Total pressure exerted by gases in container= 6 atm = p'

Partial pressure of gas A = $p_a =p= 4 atm$

Partial pressure of gas B = $p_b = 6 atm - 4 atm= 2 atm$

$p_a=p'\times \chi_A=p'\times \frac{n_a}{n_a+n_b}$..(1)

(Dalton's law of partial pressure)

$p_b=p'\times \chi_B=p'\times \frac{n_b}{n_a+n_b}$..(2)

Dividing (1) and (2):

$\frac{p_a}{p_b}=\frac{p'\times \frac{n_a}{n_a+n_b}}{p'\times \frac{n_b}{n_a+n_b}}$

$\frac{p_a}{p_b}=\frac{n_a}{n_b}=\frac{1 g \times M_b}{2 g\times M_a}$

$\frac{4 atm}{2 atm}=\frac{M_b}{2\times M_a}$

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

$4\times M_a=M_b$

The ratio of  molecular weight of A and B is 1:4.

Answer 2

Answer:

option 4

Explanation:

see attachment

hope it helps