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