Question

The concentration of \(\mathrm{N}_{2}\) in the ocean at \(25^{\circ} \mathrm{C}\) is \(445 \mu \mathrm{M} .\) The Henry's law constant for \(\mathrm{N}_{2}\) is \(0.61 \times 10^{-3} \mathrm{mol} \mathrm{L}^{-1} \mathrm{atm}^{-1} .\) Calculate the mass of \(\mathrm{N}_{2}\) in a liter of ocean water. Calculate the partial pressure of \(\mathrm{N}_{2}\) in the atmosphere.

Short answer

There are approximately 0.01246 grams of \(\mathrm{N}_{2}\) in a liter of ocean water and the partial pressure of \(\mathrm{N}_{2}\) is about 0.000271 atm.

Step-by-step solution

Calculate the Molar Mass of N2

First, based on the given concentration of \(\mathrm{N}_{2}\), we need to calculate how many moles of \(\mathrm{N}_{2}\) are in one liter of sea water. The given concentration is \(445 \mu \mathrm{M}\), which equals \(445 \times 10^{-6} \mathrm{mol/L}\). This means there are \(445 \times 10^{-6} \mathrm{mol}\) of \(\mathrm{N}_{2}\) per liter of ocean water.

Determine the Mass of N2

Once we know the concentration in moles, we can find out the mass of \(\mathrm{N}_{2}\) per liter. Given the molar mass of \(\mathrm{N}_{2}\) is 28 grams per mole, we can multiply this value by the number of moles from the first step: \(445 \times 10^{-6} \mathrm{moles} \times 28 \mathrm{g/mole} = 0.01246 \mathrm{g}\). So, there are approximately 0.01246 g of \(\mathrm{N}_{2}\) in a liter of ocean water.

Calculate the Partial Pressure of N2

To find the partial pressure of \(\mathrm{N}_{2}\), we can use the equation of Henry's law: \(P = C \times K\), where \(P\) is the partial pressure, \(C\) is the concentration, and \(K\) is Henry's law constant. We substitute the given values into the equation: \(P = 445 \times 10^{-6} \mathrm{mol/L} \times 0.61 \times 10^{-3} \mathrm{ atm/L}\). Upon solving, we find that the partial pressure of \(\mathrm{N}_{2}\) is approximately \(0.000271 \mathrm{atm}\).

Key concepts

Concentration of Gases in Solutions

Understanding the concentration of gases in solutions is crucial when studying how gases dissolve in liquids like water. A concentration tells us how much of a substance is present within a certain volume of solution. In this particular case, we're looking at nitrogen gas, \(\mathrm{N}_{2}\), in ocean water.

For gases, their concentration in a liquid solution can be described using units like \(\mu\)M, which stands for micromoles per liter. This means that if the concentration of \(\mathrm{N}_{2}\) is 445 \(\mu\)M, there are 445 micromoles of nitrogen gas in every liter of sea water. To convert micromoles into moles, multiply by \(10^{-6}\).

When calculating other properties such as the mass of the gas or its partial pressure, knowing the concentration is the first important step. It's like knowing how much chocolate chips we have before baking a batch of cookies; it sets the stage for all following calculations.

Molar Mass Calculations

Calculating the molar mass allows us to find out the mass of a substance from its concentration. Molar mass is the "weight" of one mole of a substance and is expressed in grams per mole (g/mol). For nitrogen gas, \(\mathrm{N}_{2}\), the molar mass is 28 g/mol.

Knowing the concentration of \(\mathrm{N}_{2}\) in moles per liter, we multiply it by its molar mass to find the mass of nitrogen in a liter of ocean water. We have \(445 \times 10^{-6}\) moles of nitrogen per liter. Multiplying by 28 g/mol, we find the mass as approximately 0.01246 grams.

This calculation is essential as it transforms abstract molar concentrations into tangible, real-world measurements. By converting moles into grams, we better understand how much of a substance is actually present in a given situation. This way, you know exactly how many grams of \(\mathrm{N}_{2}\) are swimming around in a liter of ocean.

Partial Pressure

Partial pressure is the pressure each gas in a mixture would exert if it alone occupied the entire volume of the container. In terms of gases in liquid solutions, this can be described using Henry’s Law. This law shows the relationship between the concentration of the gas in solution and its partial pressure in the atmosphere.

Mathematically, Henry’s law can be expressed as \(P = C \times K\), where \(P\) is the partial pressure, \(C\) is the concentration (in this case, the molar concentration of the gas), and \(K\) is the Henry's law constant.

For our \(\mathrm{N}_{2}\) example, we plug in the values: with a concentration of \(445 \times 10^{-6} \,\mathrm{mol/L}\) and Henry's law constant \(0.61 \times 10^{-3} \,\mathrm{mol/L \, atm}\), the partial pressure comes out to be \(0.000271\) atm.

This means \(\mathrm{N}_{2}\) contributes that tiny amount to the total atmospheric pressure. Understanding partial pressure is essential for everything from weather forecasting to understanding how gases behave underwater.