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Two gases A and B having molecular weights 60 and 45 respectively are enclosed in a vessel . The weight of A is 0.50g and that of B is 0.2g. The total pressure of the mixture is 750mm. Calculate partial pressure of the two gases.

$\begin{array}{1,1}(a)\;489.23mm,260.77mm\\(b)\;543.1mm,432.1mm\\(c)\;249.33mm,146.21mm\\(d)\;211.3mm,423.1mm \end {array}$

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Given
Weight of gas A = 0.50 g
Molecular weight of gas A = 60
Weight of gas B = 0.2 g
Molecular weight of gas B = 45
Pm = 750mm
From Daltons's law of partial pressure
$P'_A = P_m\times$ mole fraction of A
$ = 750\times{\large\frac{\frac{0.5}{60}}{(\large\frac{0.5}{60})+(\large\frac{0.2}{45})}}$
$ = 489.23 mm$
Now $P_m = P'_A +P'_B$
$P'_B = P_m - P'_A $
=750 - 489.23 = 260.77 mm
Hence answer is (a)
answered Feb 14, 2014 by sharmaaparna1
 

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