Question

in a mixture of helium and chlorine, occupying a volume of 12.8 l at 605.6 mmhg and 21.6 oc, it is found that the partial pressure of chlorine is 143 mmhg. what is the total mass of the sample?

Answer 1

Answer:

Mass of sample = 8.483 g

Explanation:

Given data;

Volume of mixture = 12.8 L

Pressure = 605.6 mmHg    ( 605.6 / 760 = 0.797 atm)

Temperature = 21.6 °C   (21.6 + 271.15 = 294.8 K)

Partial pressure of chlorine = 143 mmHg    ( 143/760 = 0.19 atm)

Solution:

First of all we will determine the number of moles of mixture.

PV = nRT

n = PV/RT

n = 0.797atm × 12.8L / 0.0821 atm. dm³ mol⁻¹ K⁻¹ ×294.8 K

n = 10.202 / 24.2031

n = 0.422 mol

partial pressure of chlorine is 0.19 atm so mole fraction is,

mole fraction = 0.19/0.797

mole fraction = 0.24

moles of chlorine = 0.24 × 0.422 = 0.1013 mol

moles of helium = moles of mixture - moles of chlorine

moles of helium = 0.422 - 0.1013

moles of helium = 0.3207 mol

Mass of chlorine = moles × molar mass

Mass of chlorine = 0.1013 mol × 71 g/mol

Mass of chlorine = 7.2 g

Mass of helium = moles × molar mass

Mass of helium = 0.3207 mol × 4 g/mol

Mass of helium = 1.283 g

Mass of sample = mass of chlorine + mass of helium

Mass of sample = 7.2 g + 1.283 g

Mass of sample = 8.483 g