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

B hope it helps you.

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☆What is pressure?

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☆Pressure is defined to be the amount of force exerted per area. P = F A {\Large P=\dfrac{F}{A}} P=AF. So to create a large amount of pressure, you can either exert a large force or exert a force over a small area (or do both).

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•A simple example of pressure may be seen by holding a knife to a piece of fruit. If you hold the flat part of the knife against the fruit, it won't cut the surface. The force is spread out of a large area (low pressure).

☆Hope It Helps uh!☆

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Answer 2

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.