Question

A gaseous mixture of three gases A, B and C has a pressure of 10 atms. The total number of moles of all the gases is 10. If the partial pressures of A and B are 3.0 and 1.0 atm respectively and if C has mol. wt. of 2.0, what is the weight of C in grams present in the mixture?

• A. 6
• B. 3
• C. 12
• D. 8

Answer 1

Answer: (C)

Partial pressure of A = 3 atm Partial pressure of B = 1 atm Total pressure = 10 atm Partial pressure of $ C({{P}_{C}})=A+ $ Total pressure - (partial pressure of A + partial pressure of B) $ =10-(3+1) $ $ {{P}_{c}}=10-4=6\,\text{atm} $ Mol. Wt. of C = 2 Partial pressure of C = Mole fraction of C total pressure 6 = Mole fraction of $ C\times 10 $ Mole fraction of $ C=\frac{6}{10} $ Mole fraction of $ \text{ }\!\!\!\!\text{ C }\!\!\!\!\text{ = }\frac{\text{Moles}\,\text{of }\!\!\!\!\text{ C }\!\!\!\!\text{ }}{\text{Total}\,\text{moles}} $ $ \frac{6}{10}=\frac{\text{Moles}\,\text{of}\,\text{ }\!\!\!\!\text{ C }\!\!\!\!\text{ }}{\text{10}} $ Moles of $ C=\frac{10\times 6}{10}=6 $ Moles of $ \text{ }\!\!\!\!\text{ C }\!\!\!\!\text{ = }\frac{\text{Wt}\text{.of}\text{.}\,\text{ }\!\!\!\!\text{ C }\!\!\!\!\text{ }}{\text{Mol}\text{.wt}\,\text{of}\,\text{ }\!\!\!\!\text{ C }\!\!\!\!\text{ }} $ $ \text{6 = }\frac{\text{Wt}\text{.}\,\text{of}\,\text{ }\!\!\!\!\text{ C }\!\!\!\!\text{ }}{\text{2}} $ Wt. of $ C=6\times 2=12 $