Answer:
6.022×10²² molecules
Explanation:
Given data:
Volume of nitrogen = 224 L
Pressure = standard = 1 atm
Temperature = standard = 273 K
Number of molecules = ?
Solution:
PV = nRT
1 atm × 224 L = n × 0.0821 atm.L/mol.K × 273 K
224 atm.L = n ×22.41 atm.L/mol
n = 224 atm.L/22.41 atm.L/mol
n = 10 mol
1 mole contain 6.022×10²³ molecules
10 mol×6.022×10²³ molecules/ 1 mol
60.22×10²³ molecules
6.022×10²² molecules
