1 g of a metal nitrate produces 0.86 g of m | vedclass

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Question

(A) Let the equivalent weight of the metal be $E$ and the equivalent weight of the nitrate radical $(NO_3^-)$ be $62$ and the sulphate radical $(SO_4^

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(A) Let the equivalent weight of the metal be $E$ and the equivalent weight of the nitrate radical $(NO_3^-)$ be $62$ and the sulphate radical $(SO_4^{2-})$ be $48$ (since $96/2 = 48$).According to the law of equivalence,the number of equivalents of metal nitrate equals the number of equivalents of metal sulphate.$\frac{1}{E + 62} = \frac{0.86}{E + 48}$$E + 48 = 0.86(E + 62)$$E + 48 = 0.86E + 53.32$$E - 0.86E = 53.32 - 48$$0.14E = 5.32$$E = \frac{5.32}{0.14} = 38$Thus,the equivalent weight of the metal is $38$.

$1 \ g$ of a metal nitrate produces $0.86 \ g$ of metal sulphate. The equivalent weight of the metal will be:

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