A sample of an acid, H2X, weighing 4.5g was dissolved in water. This solution was neutralised by 50.0cm3 of aqueous sodium hydroxide containing 2mol/dm3. What is the relative molecular mass, Mr, of the acid?

Answer:

19) B

Volume of NaOH = 50 cm³ = 0.05 dm³Moles of NaOH = c x v = 2 x 0.05 = 0.1 mol

Molar ratio of NaOH : H₂X = 2 : 1

So moles of H₂X = 0.1 ÷ 2 = 0.05 mol

Mr of acid = Mass ÷ Moles

Mr = 4.5 ÷ 0.05 = 90

19) B

Volume of NaOH = 50 cm³ = 0.05 dm³Moles of NaOH = c x v = 2 x 0.05 = 0.1 mol

Molar ratio of NaOH : H₂X = 2 : 1

So moles of H₂X = 0.1 ÷ 2 = 0.05 mol

Mr of acid = Mass ÷ Moles

Mr = 4.5 ÷ 0.05 = 90