Ans
Let the initial concentration of Na2CO3 solution be N₁ and volume of solution V₁ and final concentration be N₂ and volume of solution V₂
Given,
N₁ = 2 (factor 0.98)
N₂ = 1/12 (deci-normal)
V₁ = 500 ml
V₂ = ?
We know that,
N₁ x V₁ = N2 x V2
0.98 x 2 x 500 = (1/12) x V2
V2 = 11760 ml
Volume of water to be added is (11760-500) = 11260 ml (Ans.)
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