Question

18g mixture of helium and argon occupied 30L at
1 atm pressure and 27°C. Calculate the percentage
of these gases in the mixture.​

Answer 1

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Answer 2

Answer:

given :-

m = 18g of He + Ar

p = 1 atm

T= 27°c  

to find =  % of these in the mixture.

solution=

          PV = nRT

n = no of moles .

n₁ = no of molecules of helium.

n₂= no of molecules of argon.

1× 30 = (n₁+ n₂) ˣ 0.0821 ˣ ( 273 + 27)

30   = (n₁+ n₂) ₓ 0.821 ₓ 300

(n₁+n₂) =1.219

M(He) ₓn₁ + M(Ar) ₓ n₂ = 18

4n₁+ 39.9 n₂ = 18

$\frac{4n₁+ 39.9 n₂ = 18 }{(n₁+n₂) =1.219}$

35.9 n2 = 18- 9× 1.219

n2 = 0.369

now n1 =

(n1+n2)= 1.219

put the value of n2 in given equation get answer.

n1   =  0.853