Question -The density of vapour of a substance (X) at 1 atm pressure and 500 K is 0.8 kgm3. The vapour effuses through a small hole at a rate of 45 times slower than oxygen under the same condition. What is the compressibility factor (Z) of the vapour?
Options:
A .  0.974
B .  1.35
C .  1.52
D .  1.22
Answer: Option C : C rxrO2=√MO2Mx=⟮45⟯2=32Mx⇒Mx=50 dx=0.80kgm−3 Vm=1000800×50=62.5L Z=PVmRT=1×62.50.0821×500=1.52
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