Question

Dalton's Law of Partial Pressures I (two parts) For a mixture of three gases, consisting of an equal number of moles of Xe, Kr, and CF4, at a total pressure of 4.50 atm, what is the partial pressure of Xe? 2pts (Submit Answer Tries 0/5 For a mixture of two gases, consisting of 1.00 mol of X and 1.00 mol of H2, at a total pressure is 4.50 atm, what is the partial pressure of Xe? 2pts ( Submit Answer) Tries 0/5

Answer 1

Part . 1

Total pressure = 4.50 , atm

Suppose mole of each gas = x , mole

Than , total mole of gas mixture = 3x

Mole fraction of gas "Xe" = (x)/(3x) = 1/3

And , partial pressure of gas " Xe" = mole fraction * total pressure

= (1/3)*(4.50) , atm

= 1.50 , atm

Part. 2

Total pressure = 4.50 , atm

Mole of "Xe" = 1 mole

Mole of " H2 " = 1 mole

Total mole = 1+1 = 2 mole

Mole fraction of "Xe" = (1 mole)/(2 mole) = 1/2

Partial pressure of " Xe " = ( mole fraction)*(total pressure)

= (1/2)*(4.50 ), atm

= 2.25 , atm