A1.20 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%

The answer is 65.3312