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)•When volume is given in LITRES,Molarity = Number of moles of solute ÷ Volume of solution in litres•When volume is given in millilitresMolarity = (Number of moles of solute × 1000) ÷ Volume of solution in ml•When mass % and density is given,Molarity = (Mass % × Density × 10) ÷ Molar mass of SoluteMolality: It is the number of moles of solution in 1 kg of solvent.Unit of Molality: Molal (m)•When mass of solvent is in kilograms,Molality = Moles of solute ÷ Mass of solvent in kilograms•When mass of solvent is in grams,Molality = Moles of solute × 1000 ÷ Mass of solvent in gramsRelation between Molarity, Molality and Density: = + Given question: A bottle of commercial H₂SO₄ (density = 1.787 g/mL) is labelled as 86% by mass.(a) What is the molarity of the acid?(b) What volume of the acid has to be used to make 1 litre 0.2 M H2SO4?(c) What is the molality of the acid?​Answer: Molar mass of H₂SO₄ = 98 gMass = 86 g in 100 g of solutionNo. of moles = = 86/98 = 0.88 molesDensity = 1.787 g/mL = 1.787 g/mLVolume = = 56 mL(a) Molarity = (Number of moles of solute × 1000) ÷ Volume of solution in mlM = (0.88 × 1000) ÷ 56 = 880 ÷ 56 = 15.71∴ Molarity = 15.71 M(b) We KNOW that M₁V₁ = M₂V₂15.71 × V₁ = 0.2 × 1V₁ = 0.2 ÷ 15.71V₁ = 0.0127 = 0.013 L∴ Volume required = 0.013 L(c) Molality = Moles of solute × 1000 ÷ Mass of solvent in gramsm = (0.88 × 1000) ÷ 98 = 880 ÷ 98 = 8.98 ∴ Molality = 8.98 m