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The volume of 0.1 M Ca(OH)_2required toneutralize 10 mL of 0.1 N HCI |
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Answer» Solution :`0.1 M Ca(OH)_2 = 0.2 N Ca (OH)_2` APPLYING `N_1V_1 = N_2V_2` `0.2 XX V_1 = 0.1 xx 10` ` V_1 = (0.1 xx 10)/(0.2) =5mL` |
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