Question

1.0 g of metal nitrate gave 0.86 g of metal sulphate after reaction. Calculate the equivalent weight of metal.

• A. 19
• B. 48
• C. 76
• D. 38

Answer 1

The correct option is D 38

$\frac{Weightofmetalnitrate}{equivalentweightofnitrate}=\frac{Weightofmetalsulphate}{Equivalentweightofmetalsuphate}$
$E_{wS{O}_4^{2-}}=\frac{96}{2}=48$; $E_{wNO\stackrel{\ominus }{3}}=\frac{62}{1}$

So now,

$M(N{O}_3{)}_x\to M_2(S{O}_4{)}_x$

$\frac{1}{E+62}=\frac{0.86}{E+48}$

Equivalent weight of metal $=38$

(Option D)