Earlier the concept of equivalent weights was very common and the concentrations of the solutions were expressed in terms of normalities. The convenience was that the substances reacted in the ratio of their gram equivalents . So there was no need to write balanced equation to determine the amounts of the substraces reacted. However, determination or equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions, For example, KMnO_(4) has different equivalent weight in the basic medium than inthe acidic medium. Hence, now a days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality. mole fraction etc. are also used. The equivalent weight of Cu

Earlier the concept of equivalent weights was very common and the conc

1.	Earlier the concept of equivalent weights was very common and the concentrations of the solutions were expressed in terms of normalities. The convenience was that the substances reacted in the ratio of their gram equivalents . So there was no need to write balanced equation to determine the amounts of the substraces reacted. However, determination or equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions, For example, KMnO_(4) has different equivalent weight in the basic medium than inthe acidic medium. Hence, now a days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality. mole fraction etc. are also used. The equivalent weight of Cu
Answer» will be same in CuO and `Cu_(2)O` will be double in `Cu_(2)O` than in CuO will be double in CuO than in `Cu_(2)O` DEPENDS upon whether COPPER is pure or impure. Solution :In CuO, 16 parts of oxygen combines with Cu = 63.5 parts by mass. Hence, 8 parts of oxygen will combine with Cu = 31.75 parts `therefore` EQ. wt. = 31.75 In `Cu_(2)O,` 16 parts of oxygen combine with Cu `=2xx63.5` parts by mass `therefore"8 parts of oxygen will combine with Cu"` `= 63.5 ` parts `therefore" Eq. wt. = 63.5"`