0.452 g of a metal nitrate gave 0.4378 g of its metal sulphate. Calculate the equivalent weight of the metal.

0.452 g of a metal nitrate gave 0.4378 g of its metal sulphate. Calcul

1.	0.452 g of a metal nitrate gave 0.4378 g of its metal sulphate. Calculate the equivalent weight of the metal.
Answer» SOLUTION :Mass of the metal nitrate taken = 0.452 g Mass of the metal sulphate obtained = 0.4378 g SUPPOSE, equivalent mass of metal = E Using the relation, `("Mass of metal nitrate")/("Mass of metal sulphate") = ("Equivalent WEIGHT of metal nitrate")/("Equivalent weight of metal sulphate")` `0.452/0.4378 = (E+ 62)/(E + 48)` [Eq. weight of `NO_(3)^(-) = (14 + (16 xx 3))/1 = 62` [Eq. weight of `SO_(4)^(2-) = (32 + (16 xx 4))/2 = 48` or `0.452-0.4378 xx E = (0.4378 xx 62) -(0.452 xx 48) = 5.4476` or `E=5.4476/(0.452-0.4378) = 5.4476/0.0142 = 383.6`