1)it is required to prepare a solution of NaCl of volume=100ml and concentration C1=0.2mol/L from another solution of NaCl of concentration C=2mol/L.
(i) Volume of water to be added = (V2−V1)(V2-V1)(V2−V1) ml
= V1C1C2−V1\frac{V1C1}{C2}-V1C2V1C1−V1
=100×0.22−100= \frac{100\times0.2}{2}-100=2100×0.2−100
= 900 ml
(ii) molar concentration = 1.1M
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