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A mixture containing 2.24 L of H2 and 1.12 L of D2 at STP is taken inside bulb connected to another bulb by a stopcock with a small opening. The second bulb is fully evacuated, the stopcock is opened for a certain time and then closed. After effusion the 1st bulb contain 0.1 gm of D2. Find out the number of moles of H2 in second bulb
1 mol occupies 22.4 litre of volume.% mole of H2 = 2.24/22.4 = 0.01 moleMass of H2 = 2* 0.01 = 0.02 gmSimilarly, moles of D2 = 1.12/22.4 = 0.05 moleMass of D2 = 4* 0.05 = 0.2 gm Now as 0.1 gm of H2 remains in 1st bulbHenceO.1 gm of H2 will remain in 2nd bulb and 0.2 gm of D2 in second bulb also.Weight % of H2 = 0.1/(0.1+0.2) = 0.1/0.3 = 1/3 = 33.33% Hence % of D2 = 100- 33.33 = 66.67 %
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