Question

A 40.-g sample of impure KClO3 (solubility = 7.1 g per 100 g H2O at 20° C) is contaminated with 19 percent of KCl (solubility = 25.5 g per 100 g of H2O at 20° C). Calculate the minimum quantity of 20° C water needed to dissolve all the KCl from the sample. (Assume that the solubilities are unaffected by the presence of the other compound.)

Answer 1

Answer:

29.80 grams is the minimum quantity of 20° C water needed to dissolve all the KCl from the sample.

Explanation:

Mass of impure sample of $KClO_3$ = 40 g

Percentage of KCl impurity sample = 19%

Mass of KCl impurity sample = 19% of 40 g =$\frac{19}{100}\times 40 g=7.6 g$

Solubility of KCl in 100 grams of water at 20° C = 25.5 g.

Mass of water required dissolve of 1 gram KCl = $\frac{100 g}{25.5}$

The mass of water required dissolve 7.6 g of KCl at 20°C=

$\frac{100 g}{25.5}\times 7.6 =29.80 g$

29.80 grams is the minimum quantity of 20° C water needed to dissolve all the KCl from the sample.