Question

A sample of hydrogen was collected by water displacement at 23.0°C and an atmospheric pressure of 735 mmHg. Its volume is 568 mL. After water vapor is removed, what volume would the hydrogen occupy at the same conditions of pressure and temperature? (The vapor pressure of water at 23.0°C is 21 mmHg.)
A) 509 mL
B) 539 mL
C) 552 mL
D) 568 mL
E) 585 mL

Answer 1

Answer:

$V_f=552 mL$

Explanation:

Initially:

Total pressure: 735 mmHg

Water vapor pressure: 21 mmHg

Hydrogen pressure: 714 mmHg

(This is because the total pressure is divided between both gases)

When water vapor is removed:

Total pressure: 735 mmHg

Hydrogen pressure: 735 mmHg

Assuming ideal gases:

Boyle-Mariotte's Law:

$P_i*V_i = P_f * V_f$

$V_f=\frac{P_i*V_i}{P_f}$

$V_f=\frac{714 mmHg*568 mL}{735 mmHg}$

$V_f=552 mL$