1.

A 100 mL solution of KOH contains 10 milliequivalents of KOH .Calculateits strength in normality and grams /litre .

Answer»

SOLUTION :Normality `=("no.of m..E")/("VOLUME in mL")…(Eqn.1) `
`=10/100 = 0.1 `
` :. ` strength of the solution = N/10 Again,
strength in grams/LITRE = normality `xx` eq.wt = `1/10 = 56 = 5.6` grams /litre .
`(" eq.wt of KOH" ("molecular wt")/("acidity") =56/1 =56)`


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