Question

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 \(\mathrm{g} / 100 \mathrm{~g}\) water at \(10^{\circ} \mathrm{C}^{2}\)

Short answer

Answer: The solubility of lead nitrate at 10°C is 50.0 grams per 100 grams of water.

Step-by-step solution

Calculate the amount of lead nitrate after crystallization

Before cooling, there are \(57.0 \mathrm{~g}\) of lead nitrate in the solution at \(100^{\circ} \mathrm{C}\). After cooling, \(25.0 \mathrm{g}\) of lead nitrate crystallizes out, leaving the remaining lead nitrate in solution. Let's calculate the remaining lead nitrate in the solution: $$ 57.0 \mathrm{~g} - 25.0 \mathrm{~g} = 32.0 \mathrm{~g} $$ So, there are \(32.0 \mathrm{~g}\) of lead nitrate in the solution at \(10^{\circ} \mathrm{C}\).

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

As there are \(32.0 \mathrm{~g}\) of lead nitrate remaining dissolved in \(64.0 \mathrm{~g}\) of water, we can now find the solubility of lead nitrate at \(10^{\circ} \mathrm{C}\) by finding the amount of lead nitrate per \(100 \mathrm{~g}\) of water: $$ 32.0 \mathrm{~g} \cdot \frac{100 \mathrm{~g} \text{water}}{64.0 \mathrm{~g} \text{water}} = 50.0 \mathrm{~g} \text{lead nitrate} $$ Hence, the solubility of lead nitrate in \(\mathrm{g} / 100 \mathrm{~g}\) water at \(10^{\circ} \mathrm{C}\) is \(50.0 \mathrm{~g}\).