If the equivalent weight of a metal (M) is x | vedclass

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(C) For the oxide $M_mO_n$,let the atomic weight of $M$ be $A$.The mass of $M$ in the compound is $(m 	imes A) 	ext{ g}$ and the mass of oxygen is $

Vedclass · Accepted Answer

$\frac{2nx}{m}$

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(C) For the oxide $M_mO_n$,let the atomic weight of $M$ be $A$.The mass of $M$ in the compound is $(m 	imes A) 	ext{ g}$ and the mass of oxygen is $(n 	imes 16) 	ext{ g}$.According to the law of equivalence:$\frac{	ext{Mass of } M}{	ext{Equivalent weight of } M} = \frac{	ext{Mass of } O}{	ext{Equivalent weight of } O}$Given that the equivalent weight of $M$ is $x$ and the equivalent weight of oxygen is $8$:$\frac{m 	imes A}{x} = \frac{n 	imes 16}{8}$$\frac{m 	imes A}{x} = 2n$$A = \frac{2nx}{m}$

Isotope	$\%$ Natural Abundance	Molar Mass $(g/mol)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

If the equivalent weight of a metal $(M)$ is $x$ and the formula of its oxide is $M_mO_n$,then the atomic weight of $M$ is ......

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Calculate the atomic mass (average) of chlorine using the following data:

Isotope $\%$ Natural Abundance Molar Mass $(g/mol)$

$^{35}Cl$ $75.77$ $34.9689$

$^{37}Cl$ $24.23$ $36.9659$

(in $u$)

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

In terms of energy,$1 \ amu$ is equal to:

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