Question

A 25 cm3 sample of an aqueous solution of barium hydroxide, of concentration 0.146 mol dm3 was exactly neutralized by 28.7cm3 of aqueous nitric acid, according to the unbalanced equation: Ba (OH)2 (aq) + HNO3(aq) ˗˗˗˗˗˗˗˗˃ Ba(NO3)2 (aq) + H2O (l) i. i. Calculate the amount of nitric acid in terms molarity and in terms of number of moles

ii. Differentiate molarity and number of moles.

Answer 1

Ba(OH)₂ (aq) + 2HNO₃ (aq) → Ba(NO₃)₂ (aq) + 2H₂O (l)

Moles of $Ba(OH)_2 =0.146×\frac{25}{1000}=0.00365 mol$

Moles of $HNO_3=0.00365×2=0.0073mol$

Molarity of $HNO_3= \frac{0.0073}{28.7×10^{-3}}=0.25M$

The primary difference between the two comes down to mass versus volume. The molality describes the moles of a solute in relation to the mass of a solvent, while the molarity is concerned with the moles of a solute in relation to the volume of a solution.