Question

The solubility of \(\mathrm{KCl}\) at \(20^{\circ} \mathrm{C}\) is \(34.0 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water. What is the maximum amount (in grams) of \(\mathrm{KCl}\) that will dissolve in \(50.0 \mathrm{~g}\) of water at \(20^{\circ} \mathrm{C}\) ?

Short answer

The maximum amount of \(\mathrm{KCl}\) that will dissolve in \(50.0 \mathrm{~g}\) of water at \(20^{\circ} \mathrm{C}\) is \(17.0 \mathrm{~g}\)

Step-by-step solution

Setting up the proportion

The proportion can be set up as follows: \(\frac{34.0 \mathrm{~g} \; \mathrm{KCl}}{100 \mathrm{~g} \; \mathrm{water}} = \frac{x \mathrm{~g} \; \mathrm{KCl}}{50.0 \mathrm{~g} \; \mathrm{water}}\), where x is the amount of \(\mathrm{KCl}\) that will dissolve in \(50.0 \mathrm{~g}\) of water.

Solving the proportion for x

Cross-multiply to solve for x: \(x = \frac{(34.0 \mathrm{~g} \; \mathrm{KCl}) * (50.0 \mathrm{~g} \; \mathrm{water})}{100 \mathrm{~g} \; \mathrm{water}}\)

Performing the calculation

Perform the multiplication and division to find the value of x: \(x = 17.0 \mathrm{~g}\)

Key concepts

Understanding Proportion

Proportion is a mathematical concept that involves comparing two ratios or fractions. In simple terms, it means the relationship between two quantities where the ratio of one quantity to the whole remains constant. For example, if you have a ratio of 3 apples to 2 oranges, and you want to know how many apples you would need for 6 oranges, you would set up a proportion to find the answer.
\[\frac{3 \, \text{apples}}{2 \, \text{oranges}} = \frac{x \, \text{apples}}{6 \, \text{oranges}} \
\]The proportion concept is crucial because it allows us to scale quantities up or down while keeping the relationship between them fixed. In the context of the solubility of KCl in water, we use proportion to find out how much KCl can dissolve in a different amount of water than initially given.

• Proportions maintain the equality of two ratios.
• Used to solve problems involving scaling quantities.
• Helpful in real-life scenarios requiring adjustments.

When dealing with chemical calculations, proportion is often used to relate different quantities of reactants and products.

Chemistry Calculations Simplified

Chemistry calculations might seem complex at first, but they are just logical steps to find a solution. In chemistry, solving problems often includes using mathematical tools to determine concentrations, amounts, or rates. Calculations help us understand how different components behave under certain conditions.
We particularly use arithmetic operations, such as multiplication and division, to transition from known values to unknown ones. These equations often rely on the concept of molarity, volume, or, as in our current case, solubility.

• Start by identifying the known values.
• Set up an equation based on the problem's requirements.
• Use algebraic manipulation to find the unknown variable.

In the exercise with KCl, you use multiplication to adjust the amount of water and KCl proportionally, effectively simplifying the theoretical concept of solubility in practice. This calculation ensures that the given and target conditions align perfectly.

Exploring the Solubility of KCl

Solubility refers to the ability of a substance to dissolve in a solvent, like how \(\text{KCl}\) (potassium chloride) dissolves in water. \(\text{KCl}\) is a common salt that is used in various applications, including as a substitute for table salt and in lab experiments. At a specific temperature, there's a definite amount of the solute (KCl) that can dissolve in the solvent (water) before the solution becomes saturated.
The original exercise tells us the solubility of KCl at 20°C is 34.0 g per 100 g of water. This means if you add more than 34.0 g of KCl to 100 g of water at this temperature, not all of it will dissolve.

• Solubility varies with temperature.
• Saturation is the point at which no more solute can dissolve.
• Increased temperature generally increases solubility.

By understanding these concepts, we can predict how much solute will dissolve in a specific amount of solvent, like determining 17.0 g of KCl can dissolve in 50.0 g of water effortlessly.