Dalton\'s law states that the total pressure, P total of a mixture of gases in a

Question

Dalton's law states that the total pressure, P total of a mixture of gases in a container equals the sum of the pressures of each individuals gas: P total = P1 + P2 + P3 + . . .The partial pressure of the first component, P1, is equal to the mole fraction of this component, X1, times the total pressure of the mixture P1 = X1 x P total The mole fraction, X, represents the concentraction of the component in the gas mixture, so Part A Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H6) were added to the same 10.0-L container. At 23.0 C the total pressure in the container is 4.80 atm. Calculate the partial pressure of each gas in the container. Express the pressure values numerically in atmospheres, separated by commas, Enter the partial pressure of methane first, then ethane, then propane. Part B A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 465 mmHg? Express you answer numerically in millimeters of mercury.

Explanation / Answer

Assuming no reaction between N2 and O2 in given circumstances

given that 37.8 % mass is N2

let total mass of system be 'm'

mass of N2 = 0.378m

mass of O2 = 0.622m

moles of O2 = 0.622m/32

moles of N2 = 0.378m/28

mole fraction of O2 , Xo2 = (0.622m/32)/[(0.622m/32)+(0.378m/28)] = 311/527 = 0.59

=> partial pressure of O2 = Xo2 * Ptotal = 0.59 * 465 = 274.35 mmHg

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