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 mass
MD2 = 4
MH2= 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 =0.05
Final moles of H2 in 1st bulb = 0.05/2
No. 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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