Question

$1.520g$ of the hydroxide of a metal, on ignition, gave $0.995\;g$ of the oxide. The equivalent weight of the metal is:
A: $1.520$
B: $0.995$
C: $19.00$
D: $9.00$

Answer 1

The equivalent weight ( $E$ ) of any element is equal to its atomic weight ( $M$ ) divided by the valence which it assumes in the compounds or simply the charge number. The charge number refers to the number of either protons- or hydroxide equivalents which the compound contains. So, we can write the formula as: $E = \dfrac{M}{{ch\arg e{\text{ }}number}}$

Complete step by step solution:
In the question, it is given that metal hydroxide on ignition leads to the formation of the metal oxide. This means that the hydroxide as well as oxide are being involved in the same reaction. Thus, the ratio of the molecular weight of metal hydroxide and metal oxide will be equal to the ratio of the equivalent weight of metal hydroxide and metal oxide i.e.
$\dfrac{{molecular{\text{ }}weight{\text{ }}of{\text{ }}metal{\text{ }}hydroxide}}{{molecular{\text{ }}weight{\text{ }}of{\text{ }}metal{\text{ }}oxide}} = \dfrac{{equivalent{\text{ }}weight{\text{ }}of{\text{ }}metal + equivalent{\text{ }}weight{\text{ }}of{\text{ }}O{H^ - }}}{{equivalent{\text{ }}weight{\text{ }}of{\text{ }}metal + equivalent{\text{ }}weight{\text{ }}of{\text{ }}{O^{2 - }}}}$
One equivalent of metal reacts with the one mole of hydroxide ( $OH$ ), i.e. $16 + 1 = 17g$
In the same manner, one equivalent of metal reacts with the $1/2$ mole oxygen atoms, i.e., $\dfrac{{16}}{2} = 8g$ of oxygen atoms.
Let us suppose $E$ is the equivalent weight of metal and let us assume ‘ $x$ ’ as the number of gram equivalents of metal hydroxide that has been ignited.
Therefore,
Molecular weight of metal hydroxide $= 1.520g$ (Given)
Molecular weight of metal oxide $= 0.995g$ (Given)
Equivalent weight of metal $+$ equivalent weight of $O{H^ - }{\text{ = }}x\left( {E + 17} \right)$
Equivalent weight of metal $+$ equivalent weight of ${O^{2 - }}{\text{ = }}x\left( {E + 8} \right)$
Thus, substituting the values in the aforementioned ratio, we get:
$\dfrac{{1.520}}{{0.995}} = \dfrac{{x(E + 17)}}{{x(E + 8)}}$
$\therefore E = 9.06g$
Hence, the equivalent weight of the metal is $9.06g \simeq 9.0g$ . Therefore, the correct answer is Option D.

Note:
A gram equivalent refers to the ratio of the number of grams of the substance that is present and its equivalent weight. We can say that it can also be expressed as the number of the charge elements present times the total number of moles i.e. n.