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A mixture containing 1.12 L of H2 and 1.12 L of D2 at STP is taken insidea bulb connected to another bulb by a stopcock with a small opening. The second bulb is fullt evacuated, the stopcock is opened for a certain time and then closed. The first bulb is found to comtain 0.05g og H2 . Dtermine the percentage composition by weight of the gases in the second bulb.
Dear Abhijat,
Rate of diffusion ~ 1/ molecular massMD2 = 4MH2= 2
RH2/RD2 = (2)^{1/2}
As temp and volume (of the gases in the second container) remain constant
R ~ Pressure of gases (P)
P ~ no. of moles (n) => R ~ n
nD2 = nH2/ (2)^1/2
Initial moles of H2 in 1st bulb = 1.12 / 22.4 =0.05Final moles of H2 in 1st bulb = 0.05/2No. of moles of H2 in 2nd bulb = 0.05/2 => Wt of H2 = 0.05 g
nD2 = 0.05/2*1.414 = 0.0176=> Wt of D2 = 4 * 0.0176 = 0.0707 g
% of hydrogen=41.4%
(~ means proportional to)
Best Of luck
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