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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