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1.11 g of the chloride of a metal dissolved in water were treated with an excess of silver nitrate solution. The weight of the precipitated silver chloride after washing and drying was found to be 2.87 g. Calculate the equivalent weight of the metal.

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Wt.of metal chloride = 1.11g
Eq. wt. of silver = 108
Wt. of silver chloride = 2.87 g
Eq. wt. of chloride ion = 35.5
Wt.of metal chloride/Wt.of silver chloride=Eq.wt.of metal+Eq.wt.of chloride ion/Eq.wt.of silver+Eq.wt.of chloride ion
Substituting the values in the above equation, we get,
$X+35.5=\large\frac{1.11\times 143.5}{2.87}$$=55.5$
Hence, x = 55.5 – 35.5 = 20.
Therefore, the equivalent weight of the metal is 20.
answered Jun 6, 2014