A 1.20 g sample of water is injected into an evacuated 5.00 l flask at 65°c. part of the water vaporizes and creates a pressure of 187.5 mmhg. what percentage of the water vaporized?

Assume that the water vapor is an ideal gas. So,

PV = nRT

For conversion, 760 mmHg = 101325 Pa and 1,000 L = 1 m³

(187.5 mmHg) (101325 Pa/760 mmHg) (5 L) (1 m³/1,000 L) = n (8.314 m³Pa/molK) (65+273 K)

Solving for n,

n = 0.0445 mole water

Since the molar mass of water is 18 g/mol,

Mass of water vaporized = 0.0445*18 = 0.8 g water vaporized

Hence,

Percentage of water vaporized = 0.8/1.2 * 100 = 66.7%