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Azeotropic Mixtures

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Question

A mixture of $N_{2}$ and $C O_{2}$ at one atm pressure contains 56% by weight of $N_{2}$ . Partial pressure of $N_{2}$ in the mixture will be:
a)0.56 atm
b)0.44 atm
c)0.677 atm
d)0.33 atm

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Solution

Correct option - (c) 0.677atm
Step I - Number of moles calculation
Let the total weight of mixture be = 100 gm
Percentage of $N_{2} = 56 % b y w e i g h t$
Percentage of Carbon dioxide = 100% - Percentage of nitrogen
Percentage of carbon dioxide = 100 - 56 = 44%
Number of moles of Nitrogen =
$\frac{W e i g h t o f n i t r o g e n}{M o l e c u l a r w e i g h t o f n i t r o g e n}$
$N u m b e r o f m o l e s o f n i t r o g e n = \frac{56}{28} = 2 m o l e s$
$N u m b e r o f m o l e s o f c a r b o n d i o x i d e = \frac{44}{44} = 1 m o l e$
Step II - Calculation of Partial pressure of gas
$Partialpressureofgas=Molefraction×Totalpressure$
$Partialpressureofnitrogen=Molefractionofnitrogen×totalpressure$
$T o t a l p r e s s u r e = 1 a t m$
$Molefractionofnitrogen=21+2=23$
$P a r t i a l p r e s s u r e o f N i t r o g e n = \frac{2}{3} \times 1$
$P a r t i a l p r e s s u r e o f n i t r o g e n = 0.667 a t m$

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