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 ?

PV = nRT

No. of moles (n) = n

_{1}+ n

_{2}

where

n

_{1}= No. of moles of Helium

n

_{2}= No. of moles of Argon

1 atm x 10 L = (n

_{1}+ n

_{2}) x 0.0821 x (273+25)

n

_{1}

_{ }+ n

_{2}= 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

4n

_{1}+ 39.9 n

_{2}= 5 .........................................(2)

Solving equation 1 and 2 simultaneously for the value of n

_{1}and n

_{2}as follows :

4 x (n

_{1}

_{ }+ n

_{2}= 0.4087 ) .............(3)

1 x (4n

_{1}+ 39.9 n

_{2}= 5 )................(4)

subtracting eq 3 from 4 we get

35.9 n

_{2}= 3.3652

n

_{2}= 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

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

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

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