Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 111

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

Expert verified
The molarity of bromide ion in seawater is approximately \(8.34 \times 10^{-4} \, \mathrm{mol/L}\).

Step by step solution

01

Convert the concentration of bromide ion to moles

First, we need to convert the concentration of the bromide ion, given in mg/kg, to moles. To do this, we will use the molar mass of bromide ion (Br⁻), which is approximately 79.90 g/mol. The given concentration of bromide ion is 65 mg of Br⁻ per kg of seawater. We need to convert this into grams and then to moles using the molar mass of Br⁻: \(65 \, mg = 0.065 \, g \, \text{of Br⁻}\). Now, we can convert 0.065 g of Br⁻ to moles of Br⁻ using the molar mass: \(n = \frac{0.065 \, \text{g}}{79.90 \, \text{g/mol}} = 8.1352 \times 10^{-4} \, \text{mol}\). In 1 kg of seawater, there are \(8.1352 \times 10^{-4}\) moles of Br⁻.
02

Convert the mass of seawater to volume

Next, we will convert the mass of seawater (1 kg) to its volume. We are given the density of seawater, which is 1.025 g/mL. To do this, we will use the following equation: Density = Mass / Volume We will first need to convert 1 kg of seawater to grams (1000 g), then solve for the volume in milliliters: \(1.025 \, \text{g/mL} = \frac{1000 \, \text{g}}{V_{mL}}\) Now, we solve for the volume of seawater in milliliters (V_mL): \(V_{mL} = \frac{1000 \, \text{g}}{1.025 \, \text{g/mL}} = 975.61 \, \text{mL}\). Finally, we convert the volume of seawater to liters (V_L): \(V_{L} = \frac{975.61 \, \text{mL}}{1000 \, \text{mL/L}} = 0.97561 \, \text{L}\).
03

Calculate the molarity of bromide ion

Now that we know the number of moles of Br⁻ and the volume of seawater in liters, we can calculate the molarity (M) using the following equation: Molarity (M) = Number of moles of solute / Volume of solution (L) Using the values from the previous steps, we get: \(M = \frac{8.1352 \times 10^{-4} \, \text{mol}}{0.97561 \, \text{L}} = 8.3367 \times 10^{-4} \, \text{mol/L}\). Therefore, the molarity of bromide ion in seawater is approximately \(8.34 \times 10^{-4} \, \mathrm{mol/L}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of \(1.50 \mathrm{~g}\) of lead(II) nitrate is mixed with \(125 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) sodium sulfate solution. (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) What are the concentrations of all ions that remain in solution after the reaction is complete?

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathrm{b}) \mathrm{NO}_{3}^{-},(\mathrm{c}) \mathrm{NH}_{4}^{+},(\mathrm{d}) \mathrm{S}^{2-},\) (e) \(\mathrm{SO}_{4}^{2-}\). Explain briefly. [Section \(\left.4.2\right]\)

Name the spectator ions in any reactions that may be involved when each of the following pairs of solutions are mixed. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{MgSO}_{4}(a q)\) (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{Na}_{2} \mathrm{~S}(a q)\) (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\) and \(\mathrm{CaCl}_{2}(a q)\)

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\)

A \(35.0-\mathrm{mL}\) sample of \(1.00 \mathrm{M} \mathrm{KBr}\) and a \(60.0-\mathrm{mL}\) sample of \(0.600 \mathrm{M} \mathrm{KBr}\) are mixed. The solution is then heated to evaporate water until the total volume is \(50.0 \mathrm{~mL}\). What is the molarity of the KBr in the final solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free