Question

The average concentration of bromide ion in seawater is \(65 \mathrm{mg}\) of bromide ion per \(\mathrm{kg}\) of seawater. What is the molarity of the bromide ion if the density of the seawater is \(1.025 \mathrm{~g} / \mathrm{mL}\) ?

Short answer

The molarity of the bromide ion in seawater is approximately \(8.339 \times 10^{-4} M\).

Step-by-step solution

Find the mass of bromide ion per liter of seawater.

Given, the average concentration of bromide ion is 65 mg of bromide ion per kg of seawater. To convert this to the amount of bromide ion per liter of seawater, we will first find the mass of seawater in 1 liter and then multiply it by the concentration. The density of the seawater is given as 1.025 g/mL. Convert this to kg/L: Density = \(1.025 \frac{g}{mL} \times \frac{1000g}{1kg} \times \frac{1L}{1000mL}\) = 1.025 kg/L Now, we have 65 mg of bromide ion per 1.025 kg of seawater. To find the mass of bromide in 1 liter of seawater, we will multiply the density by the concentration: Mass of bromide ion in 1L seawater = concentration × mass of seawater in 1L = 65 mg/kg × 1.025 kg/L = 66.625 mg/L

Convert mass to moles.

To find the molarity, we need the number of moles of the bromide ion in 1 liter of seawater. We have already found the mass of the bromide ion in 1L of seawater. Now convert this mass from milligrams (mg) to moles using the molar mass of the bromide ion (Br⁻). The molar mass of bromide ion (Br⁻) is 79.904 g/mol. Number of moles = \(\frac{mass}{molar\ mass}\) Convert 66.625 mg to grams: 66.625 mg = \(\frac{66.625}{1000}\)g = 0.066625 g Now, find the moles of the bromide ion: Moles of bromide ion = \(\frac{0.066625g}{79.904 g/mol}\) = 8.339 x 10⁻⁴ moles

Calculate the molarity.

Now that we have the number of moles of the bromide ion in 1L of seawater, we can find the molarity. Molarity is defined as the number of moles of solute per liter of solution. Molarity = \( \frac{moles \:of\:solute}{liters\:of\:solution} \) In this case, the moles of solute are the moles of bromide ion, which we found in step 2, and the liters of solution is 1L of seawater. Molarity of bromide ion = \(\frac{8.339 \times 10^{-4} \:moles}{1 L}\) = 8.339 x 10⁻⁴ M So, the molarity of the bromide ion in seawater is approximately 8.339 x 10⁻⁴ M.