Equivalent concept for acid Base Titration and precipitation reactions. $0.98$ g of the metal sulphate was dissolved in water and excess of barium chloride was added. The precipitated barium sulphate weighted $0.95$ g. Calculated the equivalent weight of the metal.

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Solution

By law of equivalent proportion,
$\Rightarrow \frac{E_{M} + E_{S O_{4}^{2 -}}}{E_{B a} + E_{C l}} = \frac{Mass of metal sulphate}{Mass of B a S O_{4}}$
where, $E_{M}$ = Equivalent mass of $M$
$\Rightarrow E_{S O_{4}^{2 -}} = 48 g / e q u i v$
$\Rightarrow E_{B a} = \frac{137.32}{2} = 68.66 g / e q u i v$
$\Rightarrow E_{C l} = 35.5$
$∴ \frac{E_{M} + 48}{68.66 + 35.5} = \frac{0.98}{0.95}$
$\Rightarrow E_{M} = (\frac{98}{95} \times 104.16) - 48 = 59.4$
$∴$ Equivalent mass of metal= $59.4$

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Equivalent concept for acid Base Titration and precipitation reactions.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. Calculated the equivalent weight of the metal.

By law of equivalent proportion,⇒EM+ESO2−4EBa+ECl=Mass of metal sulphateMass of BaSO4where, EM= Equivalent mass of M⇒ESO2−4=48g/equiv⇒EBa=137.322=68.66g/equiv⇒ECl=35.5∴EM+4868.66+35.5=0.980.95⇒EM=(9895×104.16)−48=59.4∴ Equivalent mass of metal= 59.4

Equivalent concept for acid Base Titration and precipitation reactions. $0.98$ g of the metal sulphate was dissolved in water and excess of barium chloride was added. The precipitated barium sulphate weighted $0.95$ g. Calculated the equivalent weight of the metal.