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Chapter 12: Problem 12

How many grams of potassium chloride, \(\mathrm{KCl}\), will it take to prepare a saturated solution in \(500.0 \mathrm{~mL}\) of boiling water? (The solubility of \(\mathrm{KCl}\) at \(100^{\circ} \mathrm{C}\) is \(56.7 \mathrm{~g} / 100.0 \mathrm{~g}\) water; assume the density of water is \(1.000 \mathrm{~g} / \mathrm{mL}\).)

Short Answer

Expert verified
To prepare a saturated solution of potassium chloride, \(\mathrm{KCl}\), in \(500.0 \mathrm{~mL}\) of boiling water, it will take \(283.5 \mathrm{~g}\) of \(\mathrm{KCl}\).

Step by step solution

01

Calculate the mass of water

First, we need to find the mass of \(500.0 \mathrm{~mL}\) of water. Since the density of water is given as \(1.000 \mathrm{~g} / \mathrm{mL}\), we can use the following formula to find the mass: Mass of water = Density × Volume Mass of water = \(1.000 \frac{\mathrm g}{\mathrm{mL}}\times 500.0 \mathrm{~mL}\) Mass of water = \(500.0 \mathrm{~g}\)
02

Calculate the mass of \(\mathrm{KCl}\) that can be dissolved in \(100.0 \mathrm{~g}\) of water

We are given the solubility of \(\mathrm{KCl}\) in \(100.0 \mathrm{~g}\) of water at \(100^{\circ} \mathrm{C}\): $56.7 \mathrm{~g} / 100.0 \mathrm{~g}$ water. This means that \(56.7 \mathrm{~g}\) of \(\mathrm{KCl}\) will dissolve completely in \(100.0 \mathrm{~g}\) of boiling water.
03

Calculate the mass of \(\mathrm{KCl}\) needed to saturate \(500.0 \mathrm{~g}\) of water

Since we know the mass of \(\mathrm{KCl}\) needed to saturate \(100.0 \mathrm{~g}\) of water, we can find out the mass needed for \(500.0 \mathrm{~g}\) of water. Let's use proportion: \(\frac{56.7 \mathrm{~g} \ \mathrm{KCl}}{100.0 \mathrm{~g} \ \mathrm{water}}= \frac{x \mathrm{~g} \ \mathrm{KCl}}{500.0 \mathrm{~g} \ \mathrm{water}}\) Next, we will solve for \(x\): \(x \mathrm{~g} \ \mathrm{KCl}=\frac{500.0\mathrm {~g} \ \mathrm{water}\times56.7 \mathrm{~g} \ \mathrm{KCl}}{100.0 \mathrm{~g}\ \mathrm{water}}\)
04

Calculate the mass of \(\mathrm{KCl}\) required

Now, we will compute the product: \(x = \frac{500.0\mathrm {~g} \ \mathrm{water}\times56.7 \mathrm{~g} \ \mathrm{KCl}}{100.0 \mathrm{~g}\ \mathrm{water}}\) \(x = 283.5 \mathrm{~g} \ \mathrm{KCl}\) So, it will take \(283.5 \mathrm{~g}\) of potassium chloride, \(\mathrm{KCl}\), to prepare a saturated solution in \(500.0 \mathrm{~mL}\) of boiling water.

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Most popular questions from this chapter

Suppose \(200.0 \mathrm{~mL}\) of a \(2.50 \mathrm{M}\) solution of sodium hydroxide is combined with \(100.0 \mathrm{~mL}\) of a \(1.50 \mathrm{M}\) solution of iron(III) nitrate. (a) Write a net ionic equation for the formation of the expected precipitate. (b) What is the theoretical yield of precipitate in grams? (c) Suppose only \(10.95 \mathrm{~g}\) of precipitate is isolated. What is the percent yield for the reaction? (d) What is the molar concentration of the excess reactant in the combined solution?

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