Answer to Question #246515 in Chemistry for ace

Question

 A flask is first evacuated so that it contains no gas at all. Then, 2.2 g of CO2 is introduced into the flask. On warming to 22 °C, the gas exerts a pressure of 318 mm Hg. What is the volume of the flask? 

Answer 1

P(CO₂) = 318 mmHg

V(CO₂) = unknown

m(CO₂) = 2.2 g

R = 62.3637 L mmHg mol^-1 K^-1

T = 22°C = (22 + 273) = 295 K

Solution:

The molar mass of CO₂ is 44 g/mol

Hence,

(2.2 g CO₂) × (1 mol CO₂ / 44 g CO₂) = 0.05 mol CO₂

n(CO₂) = 0.05 mol

The ideal gas equation can be used.

The ideal gas equation can be expressed as:

PV = nRT

Rearranging the ideal gas equation gives:

V = nRT / P

Hence,

V(CO₂) = (0.05 mol × 62.3637 L mmHg mol^-1 K^-1 × 295 K) / (318 mmHg) = 2.89 L = 2.9 L

V(CO₂) = 2.9 L

V(flask) = V(CO₂) = 2.9 L

Answer: The volume of the flask is 2.9 liters