Question

a biomolecule contains 0.24percent Mg2+ by mass the no. of magnesium ions present in 2.5 g of biomolecules is 1 NA/24 2) NA/2400 3) 100NA/24 4) NA/100

Answer 1

A biomolecule contains 0.24 percent Mg²⁺ ions by mass.

We have to find the no of magnesium ions present in 2.5 g of biomolecules.

A/C to question, the biomolecule contains 0.24 % Mg²⁺ ions by mass.

it means 0.24 g of Mg²⁺ ions is present in 100g of biomolecules.

∴  $\frac{0.24}{100}\times2.5=0.006$ g of Mg²⁺ ions will be present in 2.5 g of biomolecules.

now the mass of Magnesium ions = 0.006 g

no of moles of Magnesium ions = mass of Mg²⁺ ions/molar mass of Mg²⁺

= 0.006/24 = 2.5 ×  10⁻⁴ mol

∵ no of ions = no of moles of ions × Avogadro's number

= 2.5 × 10⁻⁴ × $N_A$

= $\frac{N_A}{4000}$  

Therefore  $\frac{N_A}{4000}$ magnesium ions present in 2.5 g of biomolecules.

Answer 2

Given:

A biomolecule contains 0.24 percent Mg2+ ions by mass.

To Find:

We have to find the number of magnesium ions present in 2.5 g of biomolecules.

Solution:

By considering the given data,

the biomolecule contains 0.24 percent by mass means that 0.24 g of Mg2+ ions are present in 100 g of biomolecules.

No. of Mg2+ ions present in 2.5 g of biomolecule is,

$\frac{0.24}{100}*2.5=0.006g$

now the mass of Mg2+ ions is 0.006 g

We know that,

no. of moles of Mg ions = mass of Mg ions / molar mass of Mg ions

$=0.006/24=2.5*10^{-4}mol$

No. of ions = no. of moles of ion × Avogadro's no.

$=2.5*10^{-4}*N_{A}$

$=\frac{N_{A} }{4000}$

Hence the no. of Mg ions present in 2.5 g of biomolecules is $\frac{N_{A} }{4000}$

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