Question

The solubility of nitrous oxide is \(0.12 \mathrm{~g} / 100 \mathrm{~g}\) water at \(20{ }^{\circ} \mathrm{C}\) and 1.00 atm. What is the partial pressure required to dissolve \(0.25 \mathrm{~g}\) of the gas in \(100 \mathrm{~g}\) of water at \(20{ }^{\circ} \mathrm{C} ?\)

Short answer

The partial pressure required is approximately 2.08 atm.

Step-by-step solution

Understanding Solubility Given Conditions

The solubility of nitrous oxide at 1.00 atm and 20°C is given as 0.12 g per 100 g of water. This means that at these conditions, we can dissolve 0.12 g of nitrous oxide in 100 g of water.

Determining Required Solubility

We need to dissolve 0.25 g of nitrous oxide in the same amount of water (100 g). This is greater than the solubility at 1.00 atm. Hence, we need to calculate the pressure for this new solubility condition using Henry's Law.

Applying Henry's Law

According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid: \[ S_1 = k_H P_1 \]\[ S_2 = k_H P_2 \]where \( S_1 \) is the initial solubility, \( P_1 \) is the initial pressure, \( S_2 \) is the new solubility, and \( P_2 \) is the partial pressure to be determined.

Calculating Henry's Law Constant

First, calculate the Henry's Law constant \( k_H \) using the given solubility at 1 atm:\[ k_H = \frac{S_1}{P_1} = \frac{0.12}{1.00} = 0.12 \text{ g/atm} \]

Solving for New Partial Pressure

Using the determined \( k_H \), calculate the necessary partial pressure to dissolve 0.25 g of nitrous oxide:\[ P_2 = \frac{S_2}{k_H} = \frac{0.25}{0.12} \approx 2.08 \text{ atm} \]. This is the pressure needed to dissolve 0.25 g of nitrous oxide in 100 g of water at 20°C.

Key concepts

Gas Solubility

Gas solubility refers to the ability of a gas to dissolve in a liquid, which in simple terms, means how much of the gas can be mixed with the liquid under certain conditions. Factors that influence the solubility include temperature and pressure. In our exercise, we are looking at nitrous oxide, a colorless gas. At a given temperature, specifically at 20°C, the solubility of this gas indicates the maximum amount that can dissolve in water before the solution becomes saturated. Imagine solubility as a sponge – the sponge can only absorb so much water before any additional water starts to pool on the surface. At 1 atm pressure, you can dissolve 0.12 g of nitrous oxide in 100 g of water. If you want to dissolve more than this, you will need to alter the conditions, such as increasing the pressure as we will discuss below.

Partial Pressure

Partial pressure is a concept from the world of gases that describes the contribution of each individual gas to the total pressure of a mixture of gases. Picture a room full of different colored balloons, each balloon representing a different gas. The pressure exerted by each type of balloon (or gas) is its partial pressure. In this exercise, when we mention this concept, we're interested in how much additional pressure is needed from nitrous oxide alone to achieve a higher solubility in water. The change in partial pressure is what allows more gas to be absorbed into the water, just like squeezing more air into a balloon. By applying Henry's Law, which we will touch on in more detail in the next section, we can determine how much we need to "squeeze" to get our desired outcome of dissolving more nitrous oxide.

Chemistry Problem-Solving

Solving chemistry problems can sometimes feel like piecing together parts of a puzzle. You need to understand the theory and how to apply it practically. In dealing with gas solubility and partial pressure, Henry's Law is our main tool. Henry's Law states that the amount of gas dissolved in a liquid is directly proportional to its partial pressure above the liquid. This means if you increase the pressure, more gas will dissolve. The exercise uses Henry's Law to solve for the new conditions needed to dissolve 0.25 g of nitrous oxide. Here are the steps simplified:

• Calculate the Henry's Law constant from given conditions: \( k_H = \frac{0.12}{1.00} = 0.12 \text{ g/atm} \).
• Apply this constant to find the new partial pressure for the desired solubility: \( P_2 = \frac{0.25}{0.12} \approx 2.08 \text{ atm} \).

By carefully following these logical steps and calculations, you can manage similar questions with confidence. Remember, always start with what you know, find the constants, then apply them to uncover the unknowns!