1st Part : Volume of 100 g of 13% H₂SO₄ solution = $${{100} \over {1.02}}$$ mL = 98.04 mL

13 g of H₂SO₄ $$\equiv$$ $${{13} \over {98}}$$ $$\equiv$$ 0.1326 mol of H₂SO₄

$$\therefore$$ 98.04 mL solution contains 0.1326 mol of H₂SO₄.

Hence, molarity of the solution = $${{0.1326} \over {98.04}} \times 1000$$ = 1.35 mol L^$$-$$1

Again, molality (m) = $${{Mass\,of\,the\,solute \div \,Molar\,mass} \over {Mass\,of\,the\,solvent}} \times 1000$$

Now, 13% solution of H₂SO₄ means that 13g of H₂SO₄ is dissolved in 87g of solvent.

Thus, molality = $${{13/98} \over {87}} \times 1000 = {{13 \times 1000} \over {87 \times 98}} = 1.52$$

$$\therefore$$ Molality of 13% H₂SO₄ solution = 1.52 (m).

2nd Part : We know, Normality = Molarity $$\times$$ basicity of the acid. Here, basicity of H₂SO₄ = 2.

$$\therefore$$ N = 1.35 $$\times$$ 2 = 2.70 For dilution, N₁V₁ = N₂V₂

$$\therefore$$ 100 $$\times$$ 2.70 N = 1.5 N $$\times$$ V₂ or, V₂ = $${{100 \times 2.70} \over {1.5}}$$ or, V₂ = 180

$$\therefore$$ The sulphuric acid sample should be diluted upto 180 mL to prepare 1.5 N solution.