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x gm of a metal gave y gm of its oxide.hence equivalent weight of the metal is a)[x/(y-x)]*8 b)[(y-x)/x]*8 c)[(x+y)/x]*8 d)(x/y)*8

x gm of a metal gave y gm of its oxide.hence equivalent weight of the metal is


a)[x/(y-x)]*8


b)[(y-x)/x]*8


c)[(x+y)/x]*8


d)(x/y)*8

Grade:12

2 Answers

AskiitianExpert Shine
10 Points
12 years ago

Hi

The weight of a substance that will combine with or replace one mole of hydrogen or one-half mole of oxygen. The equivalent weight is equal to the atomic weight divided by the valence.

The number of parts by weight of an element or compound which will combine with or replace, directly or indirectly, 1.008 parts by weight of hydrogen, 8.00 parts of oxygen, or the equivalent weight of any other element or compound. For all elements, the atomic weight is equal to the equivalent weight times a small whole number, called the valence of the element.

Now as in the given question, x gm metal gives y gm oxide, therfore, x gm of metal needs (y- x )gm of oxygen. Therefore (y-x) parts of oxygen combines with x gm of metal. Therefore using simple unitary method, 1 part of oxygen combines with x/(y-x) parts of metal.

 Hence acc to definition, 8 parts of oxygen combines with x/{(y-x)8} , hence the answer is part a.

h v
33 Points
10 years ago

thanks for the answer

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