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Q.

100 ml of 1M HCl, 200 ml of 2M HCl and 300 ml of 3M HCl are mixed with enough water to get 1M solution. The volume of water to be added is (in ml)

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a

600

b

700

c

800

d

125

answer is C.

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Detailed Solution

sample A              sample B                sample C

V1 = 100ml             V2= 200ml              V3 = 300ml

M1 = 1M                 M2= 2M                  M3 = 3M

Final molarity of the mixture after adding water = 1M

therefore M = frac{sum V_iM_i}{V_{total}}

  1 = frac{100(1)+200(2)+300(3)}{600+V_{H_2O}}

600+V_{H_2O}=1400

V_{H_2O}=800ml

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