Question

6. Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g, what is the mass of Gas A?

A) 29 g
B) 12 g
C) 7.8 g
D) 37 g

Answer 1

Answer: Option d) 37 g

Solution:

This question can be solved using the Graham's Law which states that:

Rate of effusion or diffusion of gas is indirectly proportional to the square root of its Molar Mass.

For two gases A and B, this formula can be written as:

`(r_(a) )/(r_(b) ) =\sqrt{(M_(b) )/(M_(a) ) }`

`r_(a)` = Rate of effusion of gas A

`r_(b)` = Rate of effusion of gas B

`M_(a)` = Molar mass of gas A

`M_(b)` = Molar mass of gas B

We are given that, Gas A effuses 0.68 times as fast as Gas B. This means:

`r_(a)` = 0.68 x
`r_(b)`

Using these values in the formula of Graham's law, we get:

`(0.68* r_(b) )/(r_(b) )=\sqrt{(17)/(M_(a) ) }\\\\ 0.68=\sqrt{(17)/(M_(a) ) }\\\\0.68^(2)=(17)/(M_(a) )\\\\ M_(a)=(17)/(0.68^(2) )=37`

Therefore, mass of gas A is 37 g, rounded to nearest unit.