Question

2 gram of gas A introduced in a evacuated flask at 25° Celsius the pressure of the gas is one ATM not 3 gram of another gas B is introduced in the same flask so total pressure become 1.5 ATM the ratio of molecular mass of A and B is

Answer 1

refer to the attachment

Answer 2

Answer:The ratio of molecular mass of A and B is 1:3.

Explanation:

Pressure when only 2 grams of gas A was present $p^o_A=1atm$

After adding gas B the total pressure in the flask is 1.5 atm. Now the partial pressure of gas A will 1atm.

$p=p^o_A+p^o_B=1.5=1+p^o_B$

$p^o_B=0.5atm$

According to Dalton's law of partial pressure:

$p^o_a=p\times X_a,p^o_b=p\times X_b$

p= total pressure,

$p^o_a$= partial pressure of 'a' gas.

$p^o_b$= partial pressure of 'b' gas

$X_a and X_b=$ mole fraction of gas'a' and gas'b' respectively.

Mole fraction= $\frac{n_a}{n_a+n_b}$

$n_a$= number of moles of gas 'a'.$n_a=\frac{\text{given mass}}{\text{Molecular mass}}$

$p^o_A=1atm=p\times X_A=p\times \frac{\frac{2}{M_A}}{\frac{2}{M_A}+\frac{3}{M_B}}$...(1)

$p^o_B=0.5atm=p\times X_B=p\times \frac{\frac{3}{M_B}}{\frac{2}{M_A}+\frac{3}{M_B}}$..(2)

Dividing (1) and (2) we get:

$\frac{3}{1}=\frac{M_B}{M_A}$ or

$\frac{1}{3}=\frac{M_A}{M_B}$

The ratio of molecular mass of A and B is 1:3.