A 334 mL sample of hydrogen was collected over water at 23C on day when the atmospheric pressure was 756 mm Hg. The vapor pressure of water under these conditions is 17.54 mm Hg. What volume, in liters, will the dry hydrogen ( No water vapor) occupy at STP?

0.299 L

Explanation:

We'll begin by calculating the pressure of dry gas. This can be obtained as follow:

Measured pressure = 756 mm Hg

Water vapor pressure = 17.54 mm Hg

Pressure of dry gas =?

Pressure of dry gas = (Measured pressure) – (Water vapor pressure)

Pressure of dry gas = 738.46 mmHg

Finally, we shall determine the volume occupied at STP. This can be obtained as follow:

Initial volume (V₁) = 334 mL = 334 / 1000 = 0.334 L

Initial pressure (P₁) = 738.46 mmHg

Initial temperature (T₁) = 23 °C = 23 °C + 273 = 296 K

Final pressure (P₂) = 760 mmHg

Final temperature (T₂) = 273 K

Final volume (V₂)

P₁V₁/T₁ = P₂V₂/T₂

738.46 × 0.334 / 296 = 760 × V₂/273

Cross multiply

296 × 760 × V₂ = 738.46 × 0.334 × 273

224960 × V₂ = 67334.2597

Divide both side by 224960

V₂ = 67334.2597 / 224960

V₂ = 0.299 L

Thus, the volume of the dry gas st STP is 0.299 L.