In 1 g of a metal oxide,the mass of metal pr | vedclass

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(A) Mass of metal oxide = $1 \ g$ Mass of metal = $0.68 \ g$ Mass of oxygen = $1 \ g - 0.68 \ g = 0.32 \ g$ According to the law of equivalence,the nu

Vedclass · Accepted Answer

$17$

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(A) Mass of metal oxide = $1 \ g$ Mass of metal = $0.68 \ g$ Mass of oxygen = $1 \ g - 0.68 \ g = 0.32 \ g$ According to the law of equivalence,the number of equivalents of metal equals the number of equivalents of oxygen. $\frac{	ext{Mass of metal}}{	ext{Equivalent weight of metal}} = \frac{	ext{Mass of oxygen}}{	ext{Equivalent weight of oxygen}}$ $\frac{0.68}{E} = \frac{0.32}{8}$ $E = \frac{0.68 	imes 8}{0.32} = \frac{5.44}{0.32} = 17$ Therefore,the equivalent weight of the metal is $17$.

Isotope	Molar Mass $(g \ mol^{-1})$	Abundance
$^{36}Ar$	$35.96755$	$0.337 \%$
$^{38}Ar$	$37.96272$	$0.063 \%$
$^{40}Ar$	$39.9624$	$99.600 \%$

In $1 \ g$ of a metal oxide,the mass of metal precipitated is $0.68 \ g$. What is the equivalent weight of the metal?

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