Asked by Doris
8.4 g of a mixture of potassium hydroxide and potassium chloride were made up of 1 litre of aqueous solution. 20 cm of this solution required 24.2 cm of 0.1 m nitric acid for neutralisation. calculate the percentage by mass of potassium chloride in the mixture
Answers
Answered by
DrBob222
KOH + HNO3 ==> KNO3 + H2O
Do you mean 1M HNO3 or 1m HNO3. There is a difference, you know. I will assume you meant 1 M. Also you must mean 20 cm^3 and not 20 cm.
mols HNO3 = M x L = 0.1M x 0.0242 L = 0.00242.
mols KOH = same = 0.00242
g KOH = mols KOH x molar mass KOH = ?
Then %KOH = (grams KOH/mass sample)*100 = ?
Do you mean 1M HNO3 or 1m HNO3. There is a difference, you know. I will assume you meant 1 M. Also you must mean 20 cm^3 and not 20 cm.
mols HNO3 = M x L = 0.1M x 0.0242 L = 0.00242.
mols KOH = same = 0.00242
g KOH = mols KOH x molar mass KOH = ?
Then %KOH = (grams KOH/mass sample)*100 = ?
Answered by
Mandela
8.288
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