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The normality of 10% (weight/volume) acetic acid isA. 1 NB. 10 NC. 1.66 ND. 0.83 N

Answer» Correct Answer - c
Normality `= (w)/(E.V) = (10g)/(100 mL)xx(1)/(60)`
`= (10g)/((100)/(1000)xx60) = 1.66 N`


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