need help with this qn

• Coolplay

  26 Aug 09, 19:34

  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

  26 Aug 09, 22:04

  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 = 3x2 = 6g.
• yiha093

  27 Aug 09, 19:10

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

• TenSaru

  27 Aug 09, 23:28

  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 =)