1.520 g of the hydroxide of a metal on ignit | vedclass

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Question

(D) Let the equivalent weight of the metal be $x$.The equivalent weight of the hydroxide $M(OH)_n$ is $(x + 17)$.The equivalent weight of the oxide $M

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Let the equivalent weight of the metal be $x$.The equivalent weight of the hydroxide $M(OH)_n$ is $(x + 17)$.The equivalent weight of the oxide $M_2O_n$ is $(x + 8)$.According to the law of equivalence,the ratio of the weights is equal to the ratio of their equivalent weights:$\frac{	ext{Weight of hydroxide}}{	ext{Weight of oxide}} = \frac{	ext{Equivalent weight of hydroxide}}{	ext{Equivalent weight of oxide}}$$\frac{1.520}{0.995} = \frac{x + 17}{x + 8}$$1.520(x + 8) = 0.995(x + 17)$$1.520x + 12.16 = 0.995x + 16.915$$1.520x - 0.995x = 16.915 - 12.16$$0.525x = 4.755$$x = \frac{4.755}{0.525} \approx 9.057$Rounding to the nearest integer,the equivalent weight of the metal is $9$.

$1.520 \ g$ of the hydroxide of a metal on ignition gave $0.995 \ g$ of oxide. The equivalent weight of the metal is:

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