Question

A certain mixture of helium and krypton has a mass of 10.0g and occupies a volume of 20.0L at 25.0 Celcius and 1.20 atm. What is the mass of helium in the mixture?

Answer 1

First Calculate total moles of gases present:

Given that the volume of the gas mixture is 20 .00 L

PV = nRT ⇒ n = PV / RT

n = ( 1.20 atm) (20.00 L) / (0.08206 L atm/mole K) (298 K)

n = 0.98 mol

assume that the mass of He in mixture is x and mass of Kr is y

x = mass He and y = mass Kr

Equation #1 ⇒ x + y = 10.0 g

Equation #2 ⇒ x/4.0026 g/mol + y/83.798 g/mol = 0.98 mol

Now Substitute x = 10.0 - y into the second equation and solve for y:

x/4.0026 g/mol + y/83.798 g/mol = 0.98 mol

10.0 - y /4.0026 g/mol + y/83.798 g/mol = 0.98 mol

(83.798 ) (10.0 - y) + (4.0026) (y) = (0.98) (83.798 ) (4.0026)

837.98 -83.798y + 4.0026 y = 328.70

509.28 = 79.798 y

Y = 6.38 g

Therefore;

x + y = 10.0 g

x = 10.0 g-6.38 g

= 3.62 g

x = mass He; 3.62 g and y = mass Kr ; 6.38 g