Question

A gaseous mixture contains 56 g of $ {{\text{N}}_{\text{2}}}\text{, 44 g} $ of $ C{{O}_{2}} $ and 16 g of $ C{{H}_{4}}. $ The total pressure of mixture is 720 mm of Hg. The partial pressure of methane is :

• A. 75 aim
• B. 160 atm
• C. 180 atm
• D. 215 atm

Answer 1

Answer: (C)

First of all we have to calculate the number of moles. Number of moles of $ {{N}_{2}}=\frac{56}{28}=2 $ Number of moles of $ C{{O}_{2}}=\frac{44}{44}=1 $ Number of moles of $ C{{H}_{4}}=\frac{16}{16}=1 $ $ \therefore $ Total number of moles $ =2+1+1=4 $ $ \therefore $ mole fraction of $ C{{H}_{4}}=\frac{1}{4} $ $ \therefore $ partial pressure of $ C{{H}_{4}} $ = mole fraction of $ C{{H}_{4}}\times $ total pressure $ =\frac{1}{4}\times 720=180\,\text{atm} $