# at certain mixture of helium and argon weighing 5g occupies a volume of 10 l at 25 degreec and 1 atm pressur. wha t is the composition of the mixture in mass pecentage ?

Ideal Gas Law :

PV = nRT

No. of moles (n) = n1 + n2

where
n1 = No. of moles of Helium
n2 = No. of moles of Argon

1 atm x 10 L = (n1 + n2) x 0.0821 x (273+25)

n1 + n2 =  0.4087 ..........................................(1)

No. we know
No. of moles x Molar mass = given mass
Therefore
(Molar mass x No. of moles of He ) +​(Molar mass x No. of moles of argon ) = Mass of sample

4n1 + 39.9 n2 = 5 .........................................(2)

Solving equation 1 and 2 simultaneously for the value of n1 and n2 as follows :

4 x (n1 + n2 =  0.4087 ) .............(3)
1 x (​4n1 + 39.9 n2 = 5 )................(4)

subtracting eq 3 from 4 we get

35.9 n2 = 3.3652
n2 = 0.0937 moles of Argon

Mass of argon present = 0.0937 x 39.90 = 3.7386 g

Mass of helium present = 5g - 3.73863g = 1.2613 g

Mass % of He = (1.2613/5) x100 = 25.2274 % Helium

Mass % of Ar = (3.7386/5) x 100 = 74.772 % of Argon

• 89
What are you looking for?