Amixture of carbon dioxide and oxygen gases at a total pressure of 942 mm hg contains carbon dioxide at a partial pressure of 720 mm hg. if the gas mixture contains 11.6 grams of carbon dioxide, how many grams of oxygen are present?

2.6 g of O₂ are present in the mixture

Explanation:

The molar fraction concept to solve this:

Partial pressure / Total pressure = Moles of a gas / Total moles in the mixture

720 mmHg / 942 mmHg = 0.764

This is the molar fraction of CO₂

Sum of molar fraction in mixture = 1

1 - 0.764 = 0.236

This is molar fraction of O₂

Molar mass CO₂ = 44.01 g/m

Moles CO₂ = 11.6 g / 44.01 g/m → 0.263 moles

Let's make the equation:

0.236 = x / x + 0.263

0.236 ( x + 0.263) = x

0.236x + 0.062 = x

0.062 = x - 0.236x

0.062 = 0.764x

0.062 / 0.764 = x → 0.081

These are the moles of O₂

Moles . molar mass = mass

0.081 m . 32g/m = 2.6 g

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3