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Choose the Best Answer · Q17

Q.Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape. What fraction of oxygen escapes in the time required for one-half of the hydrogen to escape? (NEET phase I)

(a) 38\dfrac{3}{8}
(b) 12\dfrac{1}{2}
(c) 18\dfrac{1}{8}
(d) 14\dfrac{1}{4}
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Step 1. By Graham's law, rH2rO2=MO2MH2=322=16=4\dfrac{r_{H_2}}{r_{O_2}}=\sqrt{\dfrac{M_{O_2}}{M_{H_2}}}=\sqrt{\dfrac{32}{2}}=\sqrt{16}=4 -- hydrogen effuses 4 times faster than oxygen under the same conditions.

Step 2. In a given time t, the amount of gas that effuses is proportional to its rate. If half (0.5 mol out of 1 mol, taking equal starting moles) of the hydrogen escapes in time t, this corresponds to an "amount effused" of 0.5 mol for H2_2 in that time. …

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