a chemical engineer places 2.12 mol Ar, 1.37 mol N2 and 0.419 mol Kr in a...

a chemical engineer places 2.12 mol Ar, 1.37 mol N2 and 0.419 mol Kr in a 6.00 L piton-cylinder assemply at 25C and 0.825 atm. what is the partial pressure of Ar

Expert Solution

According to Raoult's law ,

The partial pressure of a gas = mole fraction x total pressure of the gaseous mixture

The number of moles of Ar , n = 2.12 mol

The number of moles of N₂ , n' = 1.37 mol

The number of moles of Kr , n'' = 0.419 mol

Total number of moles , N = n + n' + n''

= 2.12+1.37+0.419

= 3.909 mol

Mole fraction of Ar , X = n / N

= 2.12 / 3.909

= 0.542

Partial pressure of Ar , pAr = X * P P = total pressure = 0.825 atm

= 0.542 * 0.825 atm

= 0.447 atm

queen_honey_blossom answered 2 years ago

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