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Chapter 12: Problem 29

A \(50-g\) sample of impure \(\mathrm{KClO}_{3}\) (solubility \(=7.1 \mathrm{~g}\) per \(100 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) ) is contaminated with 10 percent of KCl (solubility \(=25.5 \mathrm{~g}\) per \(100 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) ). Calculate the minimum quantity of \(20^{\circ} \mathrm{C}\) water needed to dissolve all the \(\mathrm{KCl}\) from the sample. How much \(\mathrm{KClO}_{3}\) will be left after this treatment? (Assume that the solubilities are unaffected by the presence of the other compound.)

Short Answer

Expert verified
The minimum quantity of water needed to dissolve all the \(\mathrm{KCl}\) is approximately 19.61 g. After this treatment, approximately 43.61 g of \(\mathrm{KClO}_{3}\) will be left.

Step by step solution

01

Calculate the amount of KCl and KClO3

First, compute the actual mass of \(\mathrm{KCl}\) and \(\mathrm{KClO}_{3}\) in the sample. This can be achieved by calculating 10% and 90% of the total mass respectively. Hence, \(Mass\ of\ \mathrm{KCl} = 10/100*50 = 5g\), and \(Mass\ of\ \mathrm{KClO}_{3} = 90/100*50 = 45g\).
02

Determine the amount of water needed to dissolve all KCl

Next, determine the minimum quantity of water needed to dissolve all the \(\mathrm{KCl}\). Using the given solubility of \(\mathrm{KCl}\) (\(25.5 \mathrm{~g}\) per \(100 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\)), calculate the amount of water needed for 5g of \(\mathrm{KCl}\). Hence, \(Water\ needed\ for\ \mathrm{KCl} = (5\ g\ of\ \mathrm{KCl} * 100\ g\ of\ \mathrm{H}_{2}O) / 25.5\ g\ of\ \mathrm{KCl} \approx 19.61g\).
03

Calculate how much \(\mathrm{KClO}_{3}\) remains undissolved

From the 19.61g of water, identify how much \(\mathrm{KClO}_{3}\) is dissolved using its solubility (\(7.1 \mathrm{~g}\) per \(100 \mathrm{~g}\) of \(\mathrm{H}_{2}O\)), then subtract that amount from the initial \(45\ g\ of\ \mathrm{KClO}_{3}\). Hence,\\(\mathrm{KClO}_{3}\ dissolved = (7.1 \mathrm{~g}\ per\ 100 \mathrm{~g}\ of\ \mathrm{H}_{2}O * 19.61\ g\ of\ \mathrm{H}_{2}O ) / 100 \approx 1.39 g\),\Remaining \(\mathrm{KClO}_{3}\ = 45\ g\ of\ \mathrm{KClO}_{3} - 1.39\ g\ of\ \mathrm{KClO}_{3} \approx 43.61 g\).

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