Question

Calculate the mass of chlorine gas in \(500 \mathrm{~mL}\) of saturated solution at \(20^{\circ} \mathrm{C}\). The solubility of \(\mathrm{Cl}_{2}\) at \(20{ }^{\circ} \mathrm{C}\) is \(0.63 \mathrm{~g} / 100 \mathrm{~mL}\).

Short answer

The mass of chlorine gas in 500 mL is 3.15 grams.

Step-by-step solution

Understand the Problem

We need to calculate the mass of chlorine gas (Cl\(_2\)) dissolved in 500 mL of a saturated chlorine solution at 20°C. We know from the problem that the solubility of Cl\(_2\) at 20°C is 0.63 g per 100 mL of solution.

Determine the Mass of Chlorine in 100 mL

According to the solubility, there are 0.63 grams of Cl\(_2\) dissolved in 100 mL of a saturated solution. This is our base reference for solubility.

Scale the Solubility to 500 mL

Since 0.63 grams of Cl\(_2\) solves in 100 mL, to find the mass of Cl\(_2\) in 500 mL, we'll multiply the given solubility by 5, because 500 mL is 5 times 100 mL.

Calculate the Mass in 500 mL

Multiply the mass of Cl\(_2\) in 100 mL by 5: \[\text{Mass of \(Cl_2\)} = 0.63 \times 5 = 3.15 \text{ grams}.\]

Verify the Calculation

Recheck the multiplication: 0.63 grams per 100 mL times 5 equals 3.15 grams. The calculation is correct by checking once more: 0.63 \(\times\) 5 = 3.15 grams.

Key concepts

Solubility

Solubility refers to the maximum amount of a substance that can dissolve in a solvent at a specified temperature and pressure to form a stable solution. It is an essential concept in solution chemistry as it tells us how much of a particular solute can be flipped into a dissolved state within a particular amount of solvent. For example, in this exercise, the solubility of chlorine gas (Cl\(_2\)) is given as 0.63 g per 100 mL of solution at 20°C. This means if we have 100 mL of water at this temperature, it can hold up to 0.63 grams of Cl\(_2\) in a dissolved form before reaching saturation.

This inherent characteristic of any substance depends on the molecular interaction between the solute and the solvent. Factors such as temperature and the nature of the solute and solvent affect solubility. An increase in temperature often increases solubility for solid solutes, yet for gases like Cl\(_2\), solubility may decrease as the temperature increases. Hence, understanding solubility is crucial for predicting how much of a substance will dissolve under specific conditions.

Saturated Solution

A saturated solution is achieved when the maximum amount of solute has been dissolved in a solvent at a given temperature and pressure. At this point, any additional solute will remain undissolved in the solution. In our context, a saturated solution of chlorine gas in water at 20°C contains exactly 0.63 g of chlorine per 100 mL of water.

In such a solution, there is a dynamic equilibrium between the dissolved chlorine molecules and those in the gaseous state. This means that the rates of dissolution and precipitation of chlorine from the solution are equal. The concept of a saturated solution is significant because it indicates the limit of solubility under given conditions. If one attempts to dissolve more chlorine than the saturated amount, the excess will not dissolve, which is useful knowledge for applications that require precise concentrations of solutions.

Calculation Steps

To solve problems involving solubility and saturated solutions, it’s essential to follow logical and clear calculation steps. Here’s how it's done in our exercise:

• Step 1: First, understand that we are calculating the mass of Cl\(_2\) in 500 mL of saturated solution based on its known solubility.
• Step 2: Recognize that the given solubility is 0.63 g per 100 mL. This means each 100 mL of solution contains 0.63 g of Cl\(_2\).
• Step 3: Since we need to scale this to 500 mL, acknowledge that 500 mL is 5 times larger than 100 mL. Therefore, we must multiply the solubility amount by 5.
• Step 4: Calculate the mass of Cl\(_2\) in 500 mL by multiplying: \(0.63 \times 5 = 3.15\) grams.
• Step 5: Finally, verify the calculation by ensuring that the scaling and multiplication were applied correctly. This double-checking step is always advisable in any calculation.

By breaking down the problem into these manageable steps, it makes it easier to tackle similar tasks and ensure the accuracy of your results.