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

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Solution

Barium Sulphate mol wt $= 233.43$

Molecular Weight of sulphate $= 32 + 16 (4) = 32 + 64 = 96$

$= \frac{96}{233.43} = 41.1$

Weight of ${S O}_{4}$ in 0.95g= $\frac{41.1 \times 0.95}{100} = 0.39$

so metal weight $= 0.98 - 0.39 = 0.59$

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

Barium Sulphate mol wt=233.43 Molecular Weight of sulphate=32+16(4)=32+64=96=96233.43=41.1Weight of SO4 in 0.95g= 41.1×0.95100=0.39so metal weight=0.98−0.39=0.59

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

Barium Sulphate mol wt $= 233.43$

Molecular Weight of sulphate $= 32 + 16 (4) = 32 + 64 = 96$

$= \frac{96}{233.43} = 41.1$

Weight of ${S O}_{4}$ in 0.95g= $\frac{41.1 \times 0.95}{100} = 0.39$

so metal weight $= 0.98 - 0.39 = 0.59$