A gaseous mixture contains 56 g of textN text2 text, 44 g of CO2 and 16 g of CH4. The total pressure of mixture is 720 mm of Hg. The partial pressure of methane is :

Question

A gaseous mixture contains 56 g of textN text2 text, 44 g of CO2 and 16 g of CH4. 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

Tardigrade Admin · Accepted Answer

First of all we have to calculate the number of moles. Number of moles of N2=(56/28)=2 Number of moles of CO2=(44/44)=1 Number of moles of CH4=(16/16)=1 ∴ Total number of moles =2+1+1=4 ∴ mole fraction of CH4=(1/4) ∴ partial pressure of CH4 = mole fraction of CH4× total pressure =(1/4)× 720=180 textatm

Q. A gaseous mixture contains 56 g of N2​, 44 g of CO2​ and 16 g of CH4​. The total pressure of mixture is 720 mm of Hg. The partial pressure of methane is :

75 aim

160 atm

180 atm

215 atm

Q. A gaseous mixture contains 56 g of $N_{2}, 44 g$ 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 :