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%
(A) 95.28% (B) 98.48% (C) 87.78% (D) 90.76%
Answers & Comments
Verified answer
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.
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;
∴ 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,
∴ 70 milli moles = 70/1000 = 0.07 moles
★ Mass of H₂SO₄ present in 0.07 moles :
It means only 6.86g of H₂SO₄ out of 7.2g involves into the neutralisation reaction.
∴ Option (A) is the correct 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. So, 7.2g of H₂SO₄ is present in the solution
The specific gravity of sodium carbonate is 1.25 g/mL.
Hence, the mass of 25 mL of solution of sodium carbonate is 1.25×25=31.25 g.
The mass of HCl used for neutralization is
1000
32.9
×109.5=3.60 g.
The number of moles of HCl in 3.60 g is
36.5
3.60
=0.0987.
2 moles of HCl will neutralize 1 mole of sodium carbonate.
The number of moles of sodium carbonate neutralized by 0.0987 moles of HCl are
2
0.987
=0.04935
Thus, 31.25 g of sodium carbonate contains 0.04935 moles.
Hence, 125 g of sodium carbonate will contain
31.25
125
×0.04935=0.1974 moles.
They will neutrlaize 0.1974 moles of sulphuric acid which corresponds to 2×0.1974=0.3948 g eq of sulphuric acid.
The volume of 0.84 N sulphuric acid required will be
0.84
0.3948
=0.470 L or 470 mL.
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