need help with this qn

Coolplay SG

A mixture of 8.0 g of a monoatomic gas X and an unknown quantity of diatomic gas Y has a volume of V m^3 at s.t.p. When 20.0g of gas X is added to the mixture, under the same condiotions, the volume of the mixture is 2V m^3. Calculate the quantity of gas Y in the mixture.

[ relative molecular mass: X= 4 Y=1 ]

thanks (:

UltimaOnline SG

Let moles of Y be n.
Find moles of X in experiment 1.
Write equation 1 in terms of and V (using molar volume = 22.4dm3).
Find moles of X in experiment 2.
Write equation 2 in terms of n and V (using molar volume = 22.4dm3).
Solve simultaneous equations for n and V.

(2+n)x22.4/1000=V <— equation 1
(5+2+n)x22.4/1000=2V or (7+n)x11.2/1000=V <— equation 2

Equation 2 – Equation 1 :
(78.4+11.2n)-(44.8+22.4n)=0
33.6=11.2n
n=3
V=0.112m3

Hence no. of moles of gas X = 3. Sample mass of gas X = 3×2 = 6g.

yiha093 SG

sorry but any other way instead of using sim.eqn ?

TenSaru SG

Maybe this works...

20g of X occupy V m^3

So 8g of X by itself occupy 0.4V m^3 (V proportionate to n)

Then gas Y occupies V-0.4V = 0.6Vm^3

At stp, Volume (in m^3) =0.0224n (n is number of moles of X)

0.4V = 0.0224(8.0/4)

V=0.112m^3

0.6V = 0.672m^3

Again, Volume=0.0224XN (N is no. of moles of Y)

0.672 = 0.0224 (m/Mr) of Y

then solve =)