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What volume of water should be added to 300 mL of 0.5 M NaOH solution so as to get a solution of 0.2 M ? |
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Answer» `overset(("Dilute"))(M_(1)V_(1))-= overset(("Conc."))(M_(2)V_(2))` `0.2xxV_(1)=0.5xx300orV_(1)=(0.5xx300)/(0.2)=750` ML `:.` Volume of WATER to be ADDED `= 750-300 = 450 mL`. |
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