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A 41-g sample of impure KClO3 (solubility = 7.1 g per 100 g H2O at 20°C) is contaminated with 13 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.)

Respuesta :

Answer:

Mass of water required to dissolve all KCl is 20.90 grams.

Explanation:

Mass of the sample = 41 g

Percentage of KCl impurity = 13%

Mass of KCl =[tex]\frac{13g}{100}\times 41 g=5.33 g[/tex]

Solubility of [tex]KCLO_3[/tex] = 7.1 grams per 100 grams of water at 20°C

Mass of [tex]KClO_3=41 g-5.33 g=35.67 g[/tex]

Solubility of [tex]KCL[/tex] = 25.5 grams per 100 grams of water at 20°C

Mass of [tex]KCl=5.33 g[/tex]

Then 1 gram of  [tex]KCl[/tex] will dissolve in = [tex]\frac{100}{25.5}[/tex] grams of water

Mass of water required to dissolve 5.33 grams of [tex]KCl[/tex]:

[tex]\frac{100}{25.5g}\times 5.33=20.90 g[/tex]