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Calculate the molarity `(M)` and normality `(N)` of a solution of oxalic acid `[(COOH)_(2) . 2H_(2) O]` containing `12.6 g` of the acid in `500 mL` of the solution. |
Answer» Molar mass of `(COOH)_(2). 2H_(2)O = 2(12 + 32 + 1) + 2 (18)` `= 90 + 36 = 126 g` Equivalent mass of `(COOH)_(2). 2H_(2)O = ("Molar mass")/(n)` `= (126)/(2) = 63 g` (`n = 2`, since two `H^(o+)` inos are replaceable, i.e., dibasic acid) `M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol)("in" mL)) = (12.6 xx 1000)/(126 xx 500) = 0.02 M` `N = (W_(2) xx 1000)/(Ew_(2) xx V_(sol)("in" mL)) = (12.6 xx 1000)/(63 xx 500) = 0.4 N` (Alternatively, `N = n xx M = 2 xx 0.2 = 0.4 N)` |
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