A mixture of three gases A, B and C has a pressure of 10 atm. The total number of moles of all the gases is 10. If the partial pressure of A and B are 3 and 1 atm respectively and if C has molecular weight of 4.0, what is the weight of C in grams present in the mixture?

Total pressure of mixture of gases A, B & C  $=10atm$
Partial pressure of  

$A=(\text{No. of moles of }A\times 10)/10\text{ or }(\text{No. of moles of }A\times 10)/10=3$
Hence, No. of moles of  $A=3$
Partial pressure of  

$B=(\text{No. of moles of }B\times 10)/10(\text{since partial pressure of }B=1atm)$
No. of moles of $B=1$ . Now the No. of moles of $C=10-(3+1)=6;$ 1 mole of C weighs  $=2gm.$