Question

4 mL of concentrated H2SO4 having a density of 1.8 g / mL was diluted to one litre. 25.0 mL of this dilute solution needed 35.0 mL of an N/10 Na2CO3 solution for complete neutralization. The percentage purity of H2SO4 is

(A) 95.28% (B) 98.48% (C) 87.78% (D) 90.76%

Answer 1

Answer :

First of all we need to find mass of H₂SO₄. We know that density is measured as the ratio of mass to the volume.

$:\implies\sf\:Density=\dfrac{Mass}{Volume}$

$:\implies\sf\:1.8=\dfrac{Mass}{4}$

$:\implies\sf\:Mass=1.8\times 4$

$:\implies\bf\:Mass=7.2\:g$

So, 7.2g of H₂SO₄ is present in the solution.

After that 7.2g of H₂SO₄ is diluted to one litre.

ATQ, 35mL of N/10 Na₂CO₃ is required to neutralise 25mL of H₂SO₄ solution.

As per milli equivalent concept;

$:\implies\sf\:m_{eq}\:of\:H_2SO_4=m_{eq}\:of\:Na_2CO_3$

$:\implies\sf\:milli\:moles\times n=N\times V_{ml}$

• n factor of H₂SO₄ = 2

$:\implies\sf\:milli\:moles\times 2=\dfrac{1}{10}\times 35$

$:\implies\sf\:milli\:moles=\dfrac{35}{20}$

$:\implies\bf\:milli\:moles=1.75$

∴ No. of milli moles present in 25mL of H₂SO₄ solution = 1.75

No. of milli moles present in 1L (1000 mL) of H₂SO₄ is given by

→ 1000 × 1.75/25

→ 70 milli moles

We know that,

• 1 mole = 10³ milli moles

∴ 70 milli moles = 70/1000 = 0.07 moles

★ Mass of H₂SO₄ present in 0.07 moles :

$:\implies\sf\:Mole=\dfrac{Mass}{Molar\:mass}$

• Molar mass of H₂SO₄ = 98

$:\implies\sf\:0.07=\dfrac{Mass}{98}$

$:\implies\bf\:Mass=6.86\:g$

It means only 6.86g of H₂SO₄ out of 7.2g involves into the neutralisation reaction.

$:\implies\sf\:\%_{purity}=\dfrac{6.86}{7.2}\times 100$

$:\implies\sf\:\%_{purity}=\dfrac{686}{7.2}$

$:\implies\:\underline{\boxed{\bf{\orange{\%_{purity}=95.28\%}}}}$

∴ Option (A) is the correct answer!