How much water is to be added to dilute 10 ml of 10 N Hcl to make it decinormal?
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N1V1 = N2V2
10×10 =0.1 × V2
V2= 1000ml
Therefore, the amount of water to be added = V2 - V1 =1000- 10 = 990 ml
10×10 =0.1 × V2
V2= 1000ml
Therefore, the amount of water to be added = V2 - V1 =1000- 10 = 990 ml
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