16.25 g of metal chloride is obtained on com | vedclass

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(A) Let $E$ be the equivalent weight of the metal.The equivalent weight of oxygen $(O^{2-})$ is $8$ and the equivalent weight of chlorine $(Cl^{-})$ i

Vedclass · Accepted Answer

$18.66$

Vedclass · Answer

(A) Let $E$ be the equivalent weight of the metal.The equivalent weight of oxygen $(O^{2-})$ is $8$ and the equivalent weight of chlorine $(Cl^{-})$ is $35.5$.Using the principle of equivalence,the ratio of the weight of metal oxide to the weight of metal chloride is given by:$\frac{	ext{Weight of metal oxide}}{	ext{Weight of metal chloride}} = \frac{E + E_{O^{2-}}}{E + E_{Cl^{-}}}$Substituting the given values:$\frac{8}{16.25} = \frac{E + 8}{E + 35.5}$Cross-multiplying:$8(E + 35.5) = 16.25(E + 8)$$8E + 284 = 16.25E + 130$$284 - 130 = 16.25E - 8E$$154 = 8.25E$$E = \frac{154}{8.25} \approx 18.66$

$16.25 \ g$ of metal chloride is obtained on complete conversion of $8 \ g$ of metal oxide into metal chloride. What will be the equivalent weight of the metal?

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