Question

A metal $M$ forms the sulphate $M_{2}\left(S O_{4}\right)_{3}.$ A $0.596$ gram sample of the sulphate reacts with excess $BaCl_{2}$ to give $1.220 \, g \, BaSO_{4}.$ What is the atomic weight of $M$ (in $g/mol$ )? $\left(Molar mass of \left(BaSO\right)_{4} = 233 . 3 g / mol\right)$

• A. $26.9$
• B. $69.7$
• C. $55.8$
• D. $23$

Answer 1

Answer: (A)

The balanced equation for the given reaction is,
$ M _{2}\left( SO _{4}\right)_{3}+3 BaCl _{2} \rightarrow 2 MCl _{3}+3 BaSO _{4} $
Mole of $BaSO _{4}=\frac{1.22}{233.3} mol$
By stoichiometry of reaction:
Moles of $M _{2}\left( SO _{4}\right)_{3}=\frac{1.22}{233.3} \times \frac{1}{3}$
$=1.743 \times 10^{-3} mol$
Mass of $M _{2}\left( SO _{4}\right)_{3}=0.596 g$
$ \begin{array}{l} \therefore 1.743 \times 10^{-3}(2 M +96 \times 3)=0.596 \\ M =26.9 g / mol \end{array} $