$0.98$ g of the metal sulphate was dissolved in water and excess of barium chloride was added. The precipitated barium sulphate weighted $0.95$ g. Calculate the equivalent weight of the metal.

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Solution

Barium Sulphate molecular weight $= 233.42 g$
molecular weight of $S O_{4} = 32 + 16 (4) = 32 + 64 = 96 g$
constituent weight of $S O_{4} = \frac{96}{233.42} = 41.1$
weight of $S O_{4}$ in $0.95 g = \frac{41.1 \times 0.95}{100} = 0.39$
$\Rightarrow$ the equivalent weight of metal is $0.98 - 0.39 = 0.59$

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0.98g of the metal sulphate was dissolved in water and excess of barium chloride was added. The precipitated barium sulphate weighted 0.95g. Calculate the equivalent weight of the metal.

Barium Sulphate molecular weight =233.42 gmolecular weight of SO4=32+16(4)=32+64=96 g constituent weight of SO4=96233.42=41.1weight of SO4 in 0.95 g=41.1×0.95100=0.39⇒ the equivalent weight of metal is 0.98−0.39=0.59

$0.98$ g of the metal sulphate was dissolved in water and excess of barium chloride was added. The precipitated barium sulphate weighted $0.95$ g. Calculate the equivalent weight of the metal.