QJEEAdv78Ch7d24: JEE Advanced 1978 Chemistry - Class 11 JEEAdv78Ch7d24 Solution: What is the molarity and molality of a 13% solution (by weight) …

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.