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Question

10 moles of a mixture of gases has three gases A, B and C at a total pressure of 10 atm. The partial pressures of A and B are 3 atm and 1 atm respectively. If C has a molecular weight of 2g /mol, then the weight of C present in the mixture will be:

A
8 g
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B
12 g
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C
3 g
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D
6 g
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Solution

The correct option is B 12 g
nA+nB+nC=10 moles
PA+PB+PC=10 atm
also, PA=3 atm and, PB=1 atm
PC=10(3+1)=6 atm
nC10×10 atm=6 atm (Dalton's law)
nC=6 ωC=6 mol×2 g/mol=12 g

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