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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

The vapour density of the methyl ester of an organic monocarboxylic acid is $37$. What is the molecular weight of the acid?

The equivalent weight of $H_3PO_4$ in the following reaction is: $H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$

What are the equivalent weights of $H_3PO_4$ in the following reactions respectively?
$H_3PO_4 + OH^{-} \rightarrow H_2PO_4^{-} + H_2O$
$H_3PO_4 + 2OH^{-} \rightarrow HPO_4^{2-} + 2H_2O$
$H_3PO_4 + 3OH^{-} \rightarrow PO_4^{3-} + 3H_2O$

The equivalent weight of $Fe$ in $Fe_2O_3$ is $(Atomic \ mass \ of \ Fe = 56 \ g \ mol^{-1})$

$0.126 \ g$ of an acid is needed to completely neutralise $20 \ mL$ of $0.1 \ N$ $NaOH$ solution. The equivalent weight of the acid is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module