Question

8.4g of mixture of pottasium hydroxide and potassium chloride were made upto 1 litre of aqueous solution. 20cm3 of this solution required 24.2 cm3 of 0.1 mole of nitric acid for neutralization. calculate the percentage by mass of potassium chloride in the mixture

Answer 1

#67792 Chemistry, Other

8.4 g of mixture of potassium hydroxide and potassium chloride were made up to 1 litre of aqueous solution. 20 cm³ of this solution required 24.2 cm³ of 0.1 mole of nitric acid for neutralization. Calculate the percentage by mass of potassium chloride in the mixture?

Answer:

First of all neutralization reaction undergoes according to equation:

$$\mathrm{KOH} + \mathrm{HNO}_3 = \mathrm{KNO}_3 + \mathrm{H}_2\mathrm{O}$$ $$C_M = n/V \quad n = C_M \cdot V$$ $$n \text{ (KOH)} = n \text{ (HNO}_3\text{)} = 0.1 \cdot (24.2/1000) = 0.002 \text{ mol}$$ $$n = m/M \quad m = n \cdot M$$ $$M \text{ (KOH)} = 56 \text{ g/mol}$$ $$m \text{ (KOH)} = 0.002 \cdot 56 = 0.14 \text{ g}$$

Mass of KOH in 1 litre: $(0.14/20) \cdot 1000 = 6.8 \text{ g}$

Mass of KCl: $8.4 - 6.8 = 1.6 \text{ g}$

$$\%(\mathrm{KCl}) = (1.6/8.4) \cdot 100 = 19\%$$

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