Answer to Question #297391 in Chemistry for Jesh

Question

A 54.0 ml sample of oxygen is collected over water at 23°C and 770.0 torr pressure. What is the volume of the dry gas at STP (Vapor pressure of water at 23°C= 21.1 torr).

Answer 1

P₁ = 770.0 torr - 21.1 torr = 748.9 torr

T₁ = 23°C = 296.15 K

V₁ = 54.0 mL

STP = Standard Temperature and Pressure (T = 273.15 K and P = 760 torr)

Therefore,

P₂ = 760 torr (at STP)

T₂ = 273.15 K (at STP)

V₂ = unknown (at STP)

Solution:

The Combined Gas Law can be used:

P₁V₁/T₁ = P₂V₂/T₂

Cross-multiply to clear the fractions:

P₁V₁T₂ = P₂V₂T₁

Divide to isolate V₂:

V₂ = (P₁V₁T₂) / (P₂T₁)

Plug in the numbers and solve for V₂:

V₂ = (748.9 torr × 54.0 ml × 273.15 K) / (760 torr × 296.15 K) = 49.079 mL = 49.1 mL

V₂ = 49.1 mL

Answer: The volume of the dry gas at STP is 49.1 mL