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Q.

The density of vapour of a substance (X) at 1 atm pressure and 500 K is  0.8Kg/m3. The vapour effuses through a small hole at a rate of 4/5 times slower than oxygen under the same condition. What is the compressibility factor (z) of the vapour?
(R = 0.0821 L.atm.mol-1K-1)

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answer is 1.52.

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Detailed Solution

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rxrO2=MO2MX=(45)2=32Mx Mx=50 dx=0.80Kg/m3 Vm=1000800×50=62.5L z=PVmRT=1×62.50.0821×500=1.52

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