A) 0.33 and 0.56
B) 0.33 and 0.52
C) 0.38 and 0.52
D) 0.25 and 0.45
Correct Answer: B
Solution :
Total pressure, \[\text{P =}\frac{\text{milliequiv}\text{.of He + milliequiv}\text{.of }{{\text{O}}_{\text{2}}}}{\text{total volume}}\] \[=\frac{200\times 0.66+400\times 0.52}{400}\] \[=\frac{132+208}{400}\] \[=\frac{340}{400}=0.85\] \[\text{Partial pressure of He = }\frac{\text{millieqiv of He}}{\text{total millieqiv}}\text{ }\!\!\times\!\!\text{ p}\] \[=\frac{132}{340}\times 0.85\] = 0.33 atm Similarly, partial pressure of \[{{O}_{2}}=\frac{208}{340}\times 0.85\] = 0.52 atm
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