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1.0 g of metal nitrate gave 0.86 g of me...

1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate the equivalent weight of metal.

Text Solution

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`M (NO_(3))_(n) rarr M_(2) (SO_(4))_(n)`
`:. Eq. of M (NO_(3))_(n) = Eq. of M_(2) (SO_(4))_(n)`
`("Weight of M nitrate")/(Ew of M + Ew of NO_(3)^(ɵ)) = ("Weight. Of sulphate")/(Ew of M + Ew of SO_(4)^(2-))`
`(1.0)/(E + (62)/(1)) = (0.86)/(E + (96)/(2))`
`(1)/(E + 62) = (0.86)/(E + 48)`
`E = 38`
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