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In the crystalline solid MSO_(4).nH_(2)Oof molar mass 250 g mol^(-1), the percentage of anhydrous salt is 64 by weight. The value of n is

Answer»

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Solution :Mass of anhydrous `MSO_(4)` salt `=64/100 xx 250 = 160` g/mole
TOTAL mass of `H_(2)O` in `MSO_(4).nH_(2)O = 250-160 = 90` g/mole
Therefore, the value of N `=90/18 =5`


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