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Chapter 1: Problem 57

The solubility of lead nitrate at \(100^{\circ} \mathrm{C}\) is \(140.0 \mathrm{~g} / 100 \mathrm{~g}\) water. A solution at \(100^{\circ} \mathrm{C}\) consists of \(57.0 \mathrm{~g}\) of lead nitrate in \(64.0 \mathrm{~g}\) of water. When the solution is cooled to \(10^{\circ} \mathrm{C}, 25.0 \mathrm{~g}\) of lead nitrate crystallize out. What is the solubility of lead nitrate in \(g / 100 \mathrm{~g}\) water at \(10^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The solubility of lead nitrate at \(10^{\circ} \mathrm{C}\) is \(50 \mathrm{~g/100~g}\) water.

Step by step solution

01

Write down the given data

The solubility at \(100^{\circ} \mathrm{C}\) is \(140 \mathrm{~g/100~g}\) water, the mass of lead nitrate is \(57.0 \mathrm{~g}\), and the mass of water is \(64.0 \mathrm{~g}\). When cooled to \(10^{\circ} \mathrm{C}\), \(25.0 \mathrm{~g}\) of lead nitrate crystallize out.
02

Determine the initial mass of the solution

The initial mass of the solution at \(100^{\circ} \mathrm{C}\) is given by: $$ m_{solution} = m_{lead \thinspace nitrate} + m_{water} = 57.0 \mathrm{~g} + 64.0 \mathrm{~g} = 121.0 \mathrm{~g} $$
03

Calculate the mass of lead nitrate and water at \(10^{\circ} \mathrm{C}\)

When the solution is cooled to \(10^{\circ} \mathrm{C}\), \(25.0 \mathrm{~g}\) of lead nitrate crystallize out. So, the remaining mass of lead nitrate in the solution will be: $$ m_{lead \thinspace nitrate}^{10^{\circ}C} = m_{lead \thinspace nitrate} - m_{crystallized} = 57.0 \mathrm{~g} - 25.0 \mathrm{~g} = 32.0 \mathrm{~g} $$ The mass of water remains unchanged as it was at \(100^{\circ} \mathrm{C}\), which is \(64.0 \mathrm{~g}\).
04

Calculate the solubility of lead nitrate at \(10^{\circ} \mathrm{C}\)

Solubility is given as grams of solute per \(100 \mathrm{~g}\) of water. Thus, we can write the solubility at \(10^{\circ} \mathrm{C}\) as: $$ solubility = \frac{m_{lead \thinspace nitrate}^{10^{\circ}C}}{m_{water}} \times 100 $$ Plug in the values for the mass of lead nitrate at \(10^{\circ} \mathrm{C}\) and the mass of water: $$ solubility = \frac{32.0 \mathrm{~g}}{64.0 \mathrm{~g}} \times 100 = 50 \mathrm{~g/100~g} \thinspace water $$ So, the solubility of lead nitrate at \(10^{\circ} \mathrm{C}\) is \(50 \mathrm{~g/100~g}\) water.

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