Question

17. What is the pressure in mm of Hg , of a gas mixture that contains 1g of H2, and 8.0 of Ar in a 3.0 L container at 27°C. Ar= 31.94 18. To what temperature must

Answer 1

Molar mass of H2 = 2.016 g/mol

mass(H2)= 1 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(1 g)/(2.016 g/mol)

= 0.496 mol

Molar mass of Ar = 39.95 g/mol

mass(Ar)= 8 g

use:

number of mol of Ar,

n = mass of Ar/molar mass of Ar

=(8 g)/(39.95 g/mol)

= 0.2003 mol

Total number of mol of gas = 0.496 mol + 0.2003 mol

= 0.6963 mol

Given:

V = 3.0 L

n = 0.6963 mol

T = 27.0 oC

= (27.0+273) K

= 300 K

use:

P * V = n*R*T

P * 3 L = 0.6963 mol* 0.08206 atm.L/mol.K * 300 K

P = 5.7138 atm

= 5.7138 * 760 mmHg

= 4342 mmHg

Answer: 4.3*10^3 mmHg