# the vapour density of a mixture of gas a (molecular mass=40) and gas b (molecular mass=80) is 25. then mole% of gas b

Vikas TU
14149 Points
7 years ago
Avg.. molar mass = 25/2
M = (n1*40 + n2*80)/(n1 + n2) = 25/2
=> 80n1 + 160n2 = 25n1 + 25n2
=> 55n1 + 135n2 = 0
=> 11n1 + 27n2 = 0
n1/n2 = + 27/11..............(1)
as mol can’t be negative.
mol % of n2 = n2/(n1+n2) * 100
= 1/(n1/n2  + 1)*100
=1/(1 + 27/11) * 100
= 11/38 * 100
= 28.9%
pankaj
16 Points
4 years ago
Avg.. molar mass = 25/2
M = (n1*40 + n2*80)/(n1 + n2) = 25/2
=> 80n1 + 160n2 = 25n1 + 25n2
=> 55n1 + 135n2 = 0
=> 11n1 + 27n2 = 0
n1/n2 = + 27/11..............(1)
as mol can’t be negative.
mol % of n2 = n2/(n1+n2) * 100
= 1/(n1/n2 + 1)*100
=1/(1 + 27/11) * 100
= 11/38 * 100
= 28.9%
pankaj
16 Points
4 years ago
Avg.. molar mass = 25/2
M = (n1*40 + n2*80)/(n1 + n2) = 25/2
=> 80n1 + 160n2 = 25n1 + 25n2
=> 55n1 + 135n2 = 0
=> 11n1 + 27n2 = 0
n1/n2 = + 27/11..............(1)
as mol can’t be negative.
mol % of n2 = n2/(n1+n2) * 100
= 1/(n1/n2 + 1)*100
=1/(1 + 27/11) * 100
= 11/38 * 100
= 28.9%how is that