Answer to Question #295071 in General Chemistry for mustafa

Question

A vessel which contains 300 dm³ of nitrogen gas at 25 °C and 100.3 kPa is contaminated with carbon dioxide and water vapour. The mixture of gases is passed through concentrated sulphuric acid to remove the water vapour and a mass loss of 1.00 g is observed. Then, the mixture is bubbled through concentrated sodium hydroxide solution, a further mass loss of 16.20 g is observed. Calculate the partial pressure of each component in the mixture.

Answer 1

$PV= nRT$

Partial pressure of water is given by;

Moles= 1mole

P = (1×8.314×298.15)/0.3

P=8262.7pa

Partial pressure of CO2

Moles= 16.2/44= 0.3681moles

P= (0.3681×8.314×298.15)/0.3

P= 3041.5pa