Lawrie Ryan and Roger Norris Solutions for Chapter: Atoms, Molecules and Stoichiometry, Exercise 24: Questions

$20.0 {\mathrm{cm}}^3$ of a metal hydroxide of concentration $0.0600 \mathrm{mol} {\mathrm{dm}}^{-3}$ was titrated with $0.100 \mathrm{mol} {\mathrm{dm}}^{-3}$ hydrochloric acid. It required $24.00 {\mathrm{cm}}^3$ of the hydrochloric acid to exactly neutralise the metal hydroxide. Calculate the number of moles of metal hydroxide used.

$20.0 {\mathrm{cm}}^3$ of a metal hydroxide of concentration $0.0600 \mathrm{mol} {\mathrm{dm}}^{-3}$ was titrated with $0.100 \mathrm{mol} {\mathrm{dm}}^{-3}$ hydrochloric acid. It required $24.00 {\mathrm{cm}}^3$ of the hydrochloric acid to exactly neutralise the metal hydroxide. Calculate the number of moles of hydrochloric acid used.

$20.0 {\mathrm{cm}}^3$ of a metal hydroxide of concentration $0.0600 \mathrm{mol} {\mathrm{dm}}^{-3}$ was titrated with $0.100 \mathrm{mol} {\mathrm{dm}}^{-3}$ hydrochloric acid. It required $24.00 {\mathrm{cm}}^3$ of the hydrochloric acid to exactly neutralise the metal hydroxide. What is the simplest mole ratio of metal hydroxide to hydrochloric acid?

$20.0 {\mathrm{cm}}^3$ of a metal hydroxide of concentration $0.0600 \mathrm{mol} {\mathrm{dm}}^{-3}$ was titrated with $0.100 \mathrm{mol} {\mathrm{dm}}^{-3}$ hydrochloric acid. It required $24.00 {\mathrm{cm}}^3$ of the hydrochloric acid to exactly neutralise the metal hydroxide. Write a balanced equation for the reaction of metal hydroxide and hydrochloric acid. Use the symbol $\mathrm{M}$ for the metal. 

$0.4 \mathrm{moles}$ of aluminium sulphate, ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$, are dissolved in water and the volume of the solution was made up to $500 {\mathrm{cm}}^3$. Which one of these statements is correct?

A. The solution contains a total of $2$ moles of ions.

B. The concentration of aluminium ions is $0.8 \mathrm{mol} {\mathrm{dm}}^{-3}$.

C. The concentration of sulphate ions in solution is $1.2 \mathrm{mol} {\mathrm{dm}}^{-3}$.

D. The solution contains a total of $4$ moles of ions.