QQuestionChemistry
QuestionChemistry
2. A solution of KCl is saturated at 50°C. Use Table 9.1
(a) How many grams of solute are dissolved in 100 g of water?
(b) What is the total mass of the solution?
(c) What is the mass percent of this solution at 50°C?
(d) If the solution is heated to 100°C, how much more KCl can be dissolved in the solution without adding more water?
(e) If the solution is saturated at 100°C and then cooled to 30°C, how many grams of solute will precipitate out?
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Answer
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Step 1:I'm here to help! Before we begin, let's clarify that KCl is potassium chloride and we're looking for information in Table 9.1, which is not provided in the question.
I'll assume that Table 9.1 contains the solubility of KCl at different temperatures. Now, let's solve the problem step-by-step: (a) To find the mass of solute (KCl) dissolved in 100 g of water at 50°C, we need to look up the solubility in Table 9.1. Let's assume the solubility is X g of KCl per 100 g of water.
Step 2::
\text{Mass of KCl} = \text{Solubility (X g/100 g of water)} imes 100 \text{ g of water}
(b) To find the total mass of the solution, we need to add the mass of solute and the mass of water.
Step 3::
\text{Total mass} = \text{Mass of KCl} + \text{Mass of water}
(c) To find the mass percent of the solution, we need to divide the mass of solute by the total mass and multiply by 100.
Step 4::
\text{Mass percent} = \frac{\text{Mass of KCl}}{\text{Total mass}} imes 100
(d) When the solution is heated to 100°C, more KCl can be dissolved in the solution. The difference in solubility between 50°C and 100°C can be found in Table 9.1. Let's assume the solubility at 100°C is Y g of KCl per 100 g of water.
Step 5::
\text{Additional KCl that can be dissolved} = (\text{Y g/100 g of water} - \text{X g/100 g of water}) imes 100 \text{ g of water}
(e) When the solution is cooled from 100°C to 30°C, the solubility decreases. The difference in solubility between 100°C and 30°C can be found in Table 9.1. Let's assume the solubility at 30°C is Z g of KCl per 100 g of water.
Step 6::
\text{KCl that will precipitate} = (\text{X g/100 g of water} - \text{Z g/100 g of water}) imes 100 \text{ g of water}
Final Answer
(a) Mass of KCl: \text{Mass of KCl} = \text{Solubility (X g/ 100 g of water)} imes 100 \text{ g of water} (b) Total mass: \text{Total mass} = \text{Mass of KCl} + 100 \text{ g of water} (c) Mass percent: \text{Mass percent} = \frac{\text{Mass of KCl}}{\text{Total mass}} imes 100 (d) Additional KCl that can be dissolved: \text{Additional KCl that can be dissolved} = (\text{Y g/ 100 g of water} - \text{X g/ 100 g of water}) imes 100 \text{ g of water} (e) KCl that will precipitate: \text{KCl that will precipitate} = (\text{X g/ 100 g of water} - \text{Z g/ 100 g of water}) imes 100 \text{ g of water}
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