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If density of vapours of a substance of molar masss18gm/moleat 1 atm pressure and 500K  is 0.36 \, kg\, m^{-3}, then value of Z for the vapours is: ( TakeR=0.082\, \, L\, atm\, mole_{-1}K_{-1}  )

Option: 1

0.82


Option: 2

1.21


Option: 3

1.1


Option: 4

0.9


Answers (1)

best_answer

It is asking the Z = Vreal /Videal

fro the question M and d is given 

we can calculate volume real.

V_{real}=\frac{Molar mass}{density}=\frac{18}{0.36}

we can calculate volume ideal by PV = nRT.

V_{ideal}=\frac{nRT}{P}=\frac{1\times 0.082500}{1}

So

\\\mathrm{Z=\frac{V_{real}}{V_{ideal}}=\frac{1\times 0.082\times 500}{1}}\\\\\ \mathrm{Z=\frac{50}{0.082\times 500}=\frac{50}{41}\: =\: 1.21}

Posted by

Gunjita

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