A) 6
B) 3
C) 12
D) 8
Correct Answer: C
Solution :
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 \[C=6\times 2=12\]
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