Question

Exercise 4.
A certain mixture of helium and argon weighing 5 goce
pressure. What is the composition of the mixture in ma
pies a volume of 10 L at 25°c
mixture in mass percentage?
​

Answer 1

Answer:

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