A sample of 18.0 g of O2 has a temperature of 350 K, and pressure of 0.720 atm. What is the volume of the sample?

V = 22.36 L

Explanation:

Given data:

Mass of oxygen = 18.0 g

Temperature = 350 K

Pressure = 0.720 atm

Volume of gas = ?

Solution:

Number of moles of oxygen :

Number of moles = mass/molar mass

Number of moles = 18.0 g/ 32 g/mol

Number of moles = 0.56 mol

Volume of oxygen:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

0.720 atm × V =  0.56 mol × 0.0821 atm.L/ mol.K   × 350 K

V = 16.1 atm.L /0.720 atm

V = 22.36 L

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