Question

A mixture of neon and hydrogen gases, in a 9.30 L flask at 27 °C, contains 2.09 grams of neon and 0.637 grams of hydrogen. The partial pressure of hydrogen in the flask is _____ atm and the total pressure in the flask is _____ atm.

Answer 1

1)

Molar mass of Ne = 20.18 g/mol

Molar mass of H2 = 2.016 g/mol

n(Ne) = mass of Ne/molar mass of Ne

= 2.09/20.18

= 0.1036

n(H2) = mass of H2/molar mass of H2

= 0.637/2.016

= 0.316

n(Ne),n1 = 0.1036 mol

n(H2),n2 = 0.316 mol

Total number of mol = n1+n2

= 0.1036 + 0.316

= 0.4195 mol

2)

now find the total pressure

Given:

V = 9.30 L

n = 0.4195 mol

T = 27 oC

= (27+273) K

= 300 K

use:

P * V = n*R*T

P * 9.3 L = 0.4195 mol* 0.0821 atm.L/mol.K * 300 K

P = 1.111 atm

3)

p(H2) = n(H2)*P/total mol

= 0.316 * 1.111 / 0.4195

= 0.837 atm

p(Ne) = P - p(H2)

= 1.111 atm - 0.837 atm

= 0.274 atm

Answer:

The partial pressure of hydrogen in the flask is 0.837 atm and the total pressure in the flask is 1.111 atm.