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
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