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)" ml
= "\\frac{V1C1}{C2}-V1"
"= \\frac{100\\times0.2}{2}-100"
= 900 ml
(ii) molar concentration = 1.1M
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