100.0g of KClO 3 was added to 40.0 cm 3 of water to give a
100.0g of KClO3 was added to 40.0 cm3 of water to give a saturated solution at 298K. If the solubility of the salt is 20.0 mol dm?3 at 298K, what percentage of the salt is left undisolved? [K = 39, Cl = 35.5, O = 16]
- A) 80%
- B) 60%
- C) 5%
- D) 2%
Correct Answer: D) 2%
Explanation
To answer this question, we first need to determine the amount of KClO\(_3\) that can dissolve in 40.0 cm\(^3\) of water at 298K.
Given the solubility of KClO\(_3\) is 20.0 mol dm\(^{-3}\) at 298K, we can calculate the maximum amount of KClO\(_3\) that can dissolve in the given volume of water. To do this, we first convert the volume from cm\(^3\) to dm\(^3\):
\[\frac{40.0\, cm^3}{1000\, cm^3/dm^3} = 0.040\, dm^3\]
Next, we can find the moles of KClO\(_3\) that can dissolve in 0.040 dm\(^3\) of water:
\[(20.0\, mol\, dm^{-3})(0.040\, dm^3) = 0.8\, mol\]
Now, we can calculate the mass of dissolved KClO\(_3\) by using the given molar masses of K, Cl, and O:
\[0.8\, mol \times (39\, g/mol\, K + 35.5\, g/mol\, Cl + 3 \times 16\, g/mol\, O) = 72.0\, g\]
Since we initially had 100.0 g of KClO\(_3\), and 72.0 g dissolved, we can calculate the percentage of KClO\(_3\) left undisolved:
\[\frac{100.0\, g - 72.0\, g}{100.0\, g} \times 100\% = 28.0\%\]
However, the given answer is 2% which may be incorrect. Based on the calculations, the correct answer should be 28%. It is recommended to consult the relevant sections of the recommended textbooks for more information and clarification.