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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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600

700

800

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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