Sure! Let's walk through the solution step by step. The problem involves the solubility of \( KClO_3 \) (potassium chlorate) at two different temperatures and how much of it will precipitate out when the solution is cooled.
1. Solubility at 60°C:
At 60°C, the maximum amount of \( KClO_3 \) that can dissolve in 100 g of water is 21.5 grams.
2. Solubility at 20°C:
At 20°C, the maximum amount of \( KClO_3 \) that can dissolve in 100 g of water is 7.5 grams.
3. Amount of \( KClO_3 \) that will precipitate:
To find out how much \( KClO_3 \) will precipitate out, subtract the solubility at 20°C from the solubility at 60°C:
[tex]\[
\text{Mass of } KClO_3 \text{ precipitated} = 21.5 \text{ g} - 7.5 \text{ g} = 14.0 \text{ g}
\][/tex]
4. Rounding:
Finally, you round the result to the nearest gram. The mass of \( KClO_3 \) that will precipitate out is:
[tex]\[
14 \text{ g}
\][/tex]
Therefore, when 100 g of a saturated solution of [tex]\( KClO_3 \)[/tex] at 60°C is cooled to 20°C, 14 grams of [tex]\( KClO_3 \)[/tex] will precipitate out.
