On reduction with hydrogen,3.6 , g of an oxid | vedclass

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Question

(A) Given: Mass of metal oxide $= 3.6 \, g$,Mass of metal $= 3.2 \, g$.Mass of oxygen $= 3.6 - 3.2 = 0.4 \, g$.Equivalent weight of oxygen $= 8$.Equiv

Vedclass · Accepted Answer

$MO$

Vedclass · Answer

(A) Given: Mass of metal oxide $= 3.6 \, g$,Mass of metal $= 3.2 \, g$.Mass of oxygen $= 3.6 - 3.2 = 0.4 \, g$.Equivalent weight of oxygen $= 8$.Equivalent weight of metal $= \frac{	ext{Mass of metal}}{	ext{Mass of oxygen}} 	imes 8 = \frac{3.2}{0.4} 	imes 8 = 64$.Given equivalent weight of metal $= 32$.Valency of metal $= \frac{	ext{Equivalent weight}}{	ext{Atomic weight}} = \frac{64}{32} = 2$.Since the valency of the metal is $2$ and oxygen is $2$,the formula of the oxide is $MO$.

On reduction with hydrogen,$3.6 \, g$ of an oxide of a metal left $3.2 \, g$ of metal. If the equivalent weight of the metal is $32$,the simplest formula of the oxide would be:

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