1.

0.5 g of metal nitrate gave 0.43 g of metal sulphate

Answer»

The equivalent wt. of the metal is `0.38`
The atomic WEIGHT of the metal is 76
The atomic wt. of metal is `38x ` (X = valency of metal )
The atomic weight of metal is 19

Solution :Let metal NITRATE = `M(NO_(3))_(x)`
Metal sulphate ` = M_(2)(SO_(4))_(x)`
Eq. of `Mn(NO_(3))_(x) = " Eq. of " M_(2)(SO_(4))_(x)`
Let Eqv. Wt. of metal = `E_(M)`
`(0.5)/(E_(M) + 62) = (0.43)/(E_(M) + 96/2) `
` rArr0.5E_(M) + 0.5 xx 48 = 0.43 xx E_(M) + 62 xx 0.43`
` rArr0.5E_(M) + 24 = 0.43E_(M) + 26.66`
` RARR(0.5 - 0.43) E_(M) = 26.66 - 24`
` rArr0.07E_(M) = 2.66`
` :. E_(M) = (2.66)/(0.07) = 38`
As valence of metal is x
Atomic wt. = `38 x`


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