8.4g of a mixture of potassium hydroxide and potassium chloride were made up to 1litre of aqueous solution 20cm3 of this solution required 24.2cm3 of0.1 M nitric acid for neutralization calculate the % by mass of potassium chloride in the mixture
KCl + HNO3 ==> No reaction (but be careful here. See below**)
KOH + HNO3 ==> KNO3 + H2O
mols HNO3 used = M x L = 0.1 x 0.0242 = 0.00242
mols KOH is the same since the titration ratio is 1 mol KOH to 1 mol HNO3.
grams KOH = mols KOH x molar mass KOH = 0.00242 x 56.1 = 0.136 g in the 20 cc titrated. How much in 1 L. That is
0.136 x (1000/20) = 6.8 g KOH.
You know KCl + KOH was 8.4 so how much was KCl?
** I expect the author of the problem expects KCl + HNO3 to be no reaction and I don't believe 0.1 M HNO3 will react; however, KCl + concentrated HNO3 (that's 12 M) WILL react. KCl + HNO3 ==> KNO3 + HCl if the KCl is a solid. The HCl will be expelled as a gas.