Question

Calculate the molarity of 40% aqueous solution of sulphuric acid having density 1.02g/mL.

Answer 1

Answer- The above question is from the chapter 'Some Basic Concepts of Chemistry'.

Molarity: Number of moles of solute present in 1 litre of solution is called molarity.

Unit of Molarity = Molar (M)

•Number of moles = Given mass/Gram atomic/molecular mass

•When volume is given in litres,

Molarity = No. of moles of solute ÷ Volume of solution in litres

•When volume is given in millilitres,

Molarity = (No. of moles of solute × 1000) ÷ Volume of solution in ml

•When mass % and density is given,

Molarity = (Mass % × Density × 10) ÷ Molar mass of Solute

Question: Calculate the molarity of 40% aqueous solution of sulphuric acid having density 1.02g/mL.

Answer: Mass % = 40 %

Density = 1.02 g/mL

Mass of Solute = 2[H] + 1[S] + 4[O] = 2 + 32 + 64 = 98 g

Molarity = (Mass % × Density × 10) ÷ Molar mass of Solute

               = (40 × 102 × 10)/ 98 × 100

               = 408/98

               = 4.163 M

∴ Molarity of 40% aqueous solution of sulphuric acid having density 1.02 g/mL is 4.163 M.

Answer 2

Given Question :

Calculate the molarity of 40% aqueous solution of sulphuric acid having density 1.02g/mL.

Answer :

Given mass %=40%

Given density =1.02g/mL

Molarity of the aqueous solution =?

Molarity is the number of moles of solute present in 1 litre of the solution.

Mass of the solute (H2SO4) =2(H)+(S)+4(O)

=2×1+32+4×16

=98 gram

As we know,

Molarity =(mass%×density ×10) ÷ molar mass of solute (H2SO4)

Molarity = (40% ×102 ×10) ÷ 98 M

Molarity = 408/98 M

Molarity = 4.163 M