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Calculate the normality of `0.74 g Ca(OH)_(2)` in `10 mL` of solution. |
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Answer» Correct Answer - 2 `because "Eq.of"Ca(OH)_(2)=74//2` and volume of solution `=10xx10^(-3)L` `therefore "Normality"=(w)/(ExxV_(("in litre")))` `=(0.74xx2)/(74xx10xx10^(-3))=2N` |
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