A hydrated sulphate of metal contains $8.1$ and $43.2$ by mass of matal and sulphate respectively. The specific heat of the metal is $0.24$ cal/g. The molar mass (in g) of hydrated sulphate is:

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Solution

Suppose, hydrated sulphate of metal is $M_{2} {S O_{4}}_{n} \cdot m H_{2} O$ where n is valency of metal.
Also, $A t o m i c m a s s \times s p e c i f i c h e a t$ $=$ $6.4$
$∴$ Atomic mass of metal $= (\frac{6.4}{0.24}) = 26.67$
Now equivalent of metal $=$ equivalent of $S O_{4}^{2 -}$
$\frac{8.1}{E} = \frac{43.2}{(\frac{96}{2})}$
$∴$ Equivalent mass of metal $= 9$
$∴$ Valency of metal $= \frac{Atomic mass}{Eq. mass}$ $= \frac{26.67}{9}$ $= 2.9 = 3$ (an integer)
$∴$ Exact atomic mass of metal = $9 \times 3 = 27$ (metal is Al)
Thus molecular formula of metal sulphate is ${A l}_{2} {(S O_{4})}_{3} \cdot m H_{2} O$
Also, molar mass of ${A l}_{2} {(S O_{4})}_{3} \cdot m H_{2} O = 27 \times 2 + 96 \times 3 + 18 m = 342 + 18 m$
$∵ (342 + 18 m) g M_{2} {(S O_{4})}_{3} \cdot m H_{2} O = 18 m$
$∴ 100 g M_{2} {(S O_{4})}_{3} \cdot m H_{2} O h a s H_{2} O = 18 m \times 100 / (342 + 18 m)$
i.e., % of $H_{2} O = \frac{18 \times m \times 100}{(342 + 18 m)}$
or $(18 m \times 100) / (342 + 18 m) = (100 - 8.1 - 43.2) = 48.7$
$∴ m = 18$ ,
Thus, formula of hydrated sulphate = ${A l}_{2} {(S O_{4})}_{3} \cdot 18 H_{2} O$
And molar mass of ${A l}_{2} {(S O_{4})}_{3} \cdot 18 H_{2} O = 54 + 93 \times 3 + 18 \times 18 = 666$ g/mol.

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