Question

please solve this question​

Answer 1

Explanation:

$at \: stp \div \\ 22.4litre = 48g(ozone \: weight) \\ 1.12 \: l = 48 \div 22.4 \times 1.12 \\ 22.4 \: l = 2.4 \: g \\ \\ 22.4 \: l \: = 1mole \: \\ 1.12 \: l \: = 1 \div 22.4 \times 1.12 \\ 22.4 \: \: l \: = 0.05moles \\ it \: contains \: 2.4g \: and \: 0.5moles \: of \: \\ o3 \: for \: 1.12l \\ \\ \\ \\ hope \: its \: helpful$