Question

A 2.00 g sample of water is injected into a closed evacuated 5.00 liter flask at 65°C. What percent (by mass) of the water will be vapor when equilibrium is reached?   (the vapor pressure of water at 65°C is 187.5 mmHg)

40.0%

59.9%

88.3%

100.%

25.5%

Answer 1

calculate the mass of water in vapour phase

Given:
P = 187.5 mm Hg
= (187.5/760) atm
= 0.2467 atm
V = 5.0 L
T = 65.0 oC
= (65.0+273) K
= 338 K

find number of moles using:
P * V = n*R*T
0.2467 atm * 5 L = n * 0.08206 atm.L/mol.K * 338 K
n = 4.447*10^-2 mol

Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

use:
mass of H2O,
m = number of mol * molar mass
= 4.447*10^-2 mol * 18.016 g/mol
= 0.8012 g

mass % = mass of vapour * 100 / mass of sample
= 0.8012 * 100 / 2.00
= 40.0 %
Answer: 40.0 %