Question

If mass percentage of mg2+ in a biomolecule is 0.2%, then the number of mg2+ions present in 10g of biomolecule is (at weight of mg = 24)

Answer 1

mass percentage of Mg²+ in biomolecule is 0.2 %

so, mass of Mg²+ in 10g of biomolecule = 0.2% of 10g

= 0.2/100 × 10g

= 0.02g

now, mole of Mg²+ ions = given mass/atomic mass

= 0.02g/24g/mol

= 0.01/12 mol

= (1/1200) mol

from Avogadro's law,

1 mole of ions = 6.023 × 10²³ ions

so, (1/1200) mole of ions = 1/1200 × 6.023 × 10²³ ions

= (6.023/12) × 10²¹ ions

= 0.5019 × 10²¹ ions

≈ 5 × 10^20 ions

hence, number of Mg²+ ions present in the 10g of biomolecule is 5 × 10^20.