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Ways of Representing the Concentration of a Solution

68.4 g of A...

Question

$68.4 g$ of $A l_{2} (S O_{4})_{3}$ is dissolved in enough water to make $500 m L$ of solution. If aluminium sulphate dissociates completely then molar concentration of aluminium ions $(A l^{3 +})$ and sulphate ions $(S O_{4}^{2 -})$ in the solution are respectively (molar mass of $A l_{2} (S O_{4})_{3} = 342 g m o l^{- 1})$

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Solution

Given mass of $A l_{2} (S O_{4})_{3} = 68.4$ g
Molar nass of $A l_{2} (S O_{4})_{3} = 342 g m o l^{- 1}$
Volume $= 500$ mL $= 0.5$ L
No. of moles $= \frac{68.4}{342} = 0.2$
Molarity $= \frac{0.2}{0.5} = 0.4$
$A l_{2} (S O_{4})_{3} \to 2 A l^{+ 3} + 3 S O_{4}^{2 -}$
Molar concentration of $A l^{+ 3}$ ions $= 2 \times 0.4 = 0.8$
Molar concentration of $S O_{4}^{- 2}$ ions $= 3 \times 0.4 = 1.2$

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