Molarity Practice Quiz

Reviewed by Zohra Sattar Waxali
Zohra Sattar Waxali, PhD (Chemistry) |
Chemistry
Review Board Member
Dr. Zohra Sattar Waxali earned her doctorate in chemistry and biochemistry from Northwestern University, specializing in the metallomes of cardiac cells and stem cells, and their impact on biological function. Her research encompasses the development of arsenoplatin chemotherapeutics, stapled peptide estrogen receptor inhibitors, and antimicrobial natural products.
 , PhD (Chemistry)
By Eddy8829
E
Eddy8829
Community Contributor
Quizzes Created: 19 | Total Attempts: 61,913
| Attempts: 18,261
SettingsSettings
Please wait...
  • 1/10 Questions

    What is the molarity of 5.30 g of Na2CO3 dissolved in 400.0 mL solution?

    • 0.105M
    • 0.125M
    • 0.25M
    • 0.225M
Please wait...
About This Quiz

Are you ready to take this molarity practice quiz that we have here for you? If so, take the quiz then! The concentration of a solution can be calculated even before it is formed by the use of the number of moles they have. Do you have an upcoming chemistry exam where you need to study morality? This quiz will help you practice molarities calculations. Give it a try, and all the best! Let's go for it!

Molarity Practice Quiz - Quiz

Quiz Preview

  • 2. 

    The molarity of 10.00 g of NaOH in 1900.0 mL of solution is

    • 0.152M

    • 0.178M

    • 0.252M

    • None of these

    Correct Answer
    A. None of these
    Explanation
    Molarity is defined as the moles of solute divided by the liters of solution. The molar mass of NaOH is 40 g/mol, so this solution contains 10.00 g NaOH/ 40 g/mol = 0.25 moles of NaOH. 1900.0 mL is 1.9 L. As such, the molarity of the solution is 0.25 mol / 1.9 L = 0.13 M, which is not one of the answer choices given.

    Rate this question:

  • 3. 

    Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in seawater?

    • 0.479M

    • 0.488M

    • 0.499M

    • 0.610M

    Correct Answer
    A. 0.479M
    Explanation
    The molarity of a solution is defined as the number of moles of solute per liter of solution. In this case, the solute is sodium chloride (NaCl) and the solution is seawater. To calculate the molarity, we need to convert the mass of NaCl to moles and then divide by the volume of the solution in liters. Given that seawater contains 28.0 g of NaCl per liter, we can convert this to moles using the molar mass of NaCl (58.44 g/mol). Dividing the moles of NaCl by the volume of the solution in liters gives us the molarity. Calculating this value, we find that the molarity of sodium chloride in seawater is 0.479M.

    Rate this question:

  • 4. 

    What is the molarity of a solution made by dissolving 20.0 g of H3PO4 in 50.0 mL of solution?

    • 3.08M

    • 4.08M

    • 5.08M

    • 6.08M

    Correct Answer
    A. 4.08M
    Explanation
    The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we need to first calculate the moles of H3PO4. The molar mass of H3PO4 is 97.99 g/mol. By dividing the given mass of H3PO4 (20.0 g) by its molar mass, we find that there are 0.204 moles of H3PO4. Next, we convert the given volume of the solution (50.0 mL) to liters by dividing it by 1000. The volume of the solution is 0.050 L. Finally, we divide the moles of H3PO4 by the volume of the solution to find the molarity: 0.204 moles / 0.050 L = 4.08 M.

    Rate this question:

  • 5. 

    0.450 moles of NaCl are dissolved in 95.0 mL of water. Calculate the molarity of the NaCl solution.

    • 4.7M

    • 4.8M

    • 4.9M

    • 5.0M

    Correct Answer
    A. 4.7M
    Explanation
    To calculate the molarity of a solution, we need to divide the number of moles of solute by the volume of the solution in liters. In this case, we have 0.450 moles of NaCl dissolved in 95.0 mL of water. To convert the volume to liters, we divide by 1000, giving us 0.095 L. Dividing the moles by the volume in liters, we get a molarity of 4.737 M. Rounded to the nearest tenth, the molarity of the NaCl solution is 4.7M.

    Rate this question:

  • 6. 

    What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?

    • 0.15M

    • 0.167M

    • 0.172M

    • 0.184M

    Correct Answer
    A. 0.167M
    Explanation
    To find the molarity of a solution, we need to calculate the number of moles of solute (NaOH) and divide it by the volume of the solution in liters.

    First, we need to convert the mass of NaOH from grams to moles. The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol. So, 5.00 g of NaOH is equal to 5.00 g / 40.00 g/mol = 0.125 mol.

    Next, we need to convert the volume of the solution from milliliters to liters. 750.0 mL is equal to 750.0 mL / 1000 mL/L = 0.750 L.

    Finally, we divide the number of moles (0.125 mol) by the volume of the solution in liters (0.750 L) to get the molarity. 0.125 mol / 0.750 L = 0.1667 M, which rounds to 0.167 M.

    Therefore, the molarity of the solution is 0.167M.

    Rate this question:

  • 7. 

    15 g of NaCl occupy a volume of 75 mL. What is the molarity of the solution?

    • 3.4M

    • 3.6M

    • 3.8M

    • 4.0M

    Correct Answer
    A. 3.4M
    Explanation
    The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we are given the mass of NaCl (15g) and the volume of the solution (75 mL). We need to convert the volume from milliliters to liters by dividing it by 1000. Then, we can calculate the moles of NaCl by dividing the mass by its molar mass. Finally, we divide the moles by the volume in liters to find the molarity. The correct answer is 3.4M.

    Rate this question:

  • 8. 

    What is the molarity of sodium chloride in river water?

    • 0.455M

    • 0.488M

    • 0.510M

    • None of these

    Correct Answer
    A. None of these
    Explanation
    Waters can have varying salinities, so there is not enough information to calculate an answer to this question. Additionally, rivers are freshwater sources and likely have low concentrations of sodium chloride.

    Rate this question:

  • 9. 

    45.0 g of Ca(NO3)2 was used to create a 1.3 M solution. What is the volume of the solution?

    • 115ml

    • 190ml

    • 210ml

    • 250ml

    Correct Answer
    A. 210ml
    Explanation
    To find the volume of the solution, we can use the formula:

    Molarity (M) = moles of solute / volume of solution (L)

    Given that the molarity is 1.3 M and the moles of solute is 45.0 g of Ca(NO3)2, we need to find the volume of the solution.

    First, we need to convert the mass of Ca(NO3)2 to moles. The molar mass of Ca(NO3)2 is 164 g/mol, so:

    moles of Ca(NO3)2 = mass / molar mass
    moles of Ca(NO3)2 = 45.0 g / 164 g/mol
    moles of Ca(NO3)2 = 0.2744 mol

    Now, we can rearrange the formula to solve for the volume of the solution:

    volume of solution = moles of solute / Molarity
    volume of solution = 0.2744 mol / 1.3 mol/L
    volume of solution ≈ 0.211 L

    Since the volume is given in milliliters, we convert it to mL:

    volume of solution = 0.211 L * 1000 mL/L
    volume of solution ≈ 211 mL

    Therefore, the volume of the solution is 210 mL.

    Rate this question:

  • 10. 

    How many moles of Ca(NO3)2 are there in 75 mL of 0.25 M solution?.

    • 0.0188mol

    • 0.02mol

    • 0.0321mol

    • 0.044mol

    Correct Answer
    A. 0.0188mol
    Explanation
    The number of moles of a substance can be calculated using the formula: moles = concentration (M) × volume (L). In this case, the concentration is given as 0.25 M and the volume is given as 75 mL. However, the volume needs to be converted to liters by dividing it by 1000 (since 1 L = 1000 mL). Therefore, the volume is 75 mL ÷ 1000 = 0.075 L. Plugging these values into the formula, we get moles = 0.25 M × 0.075 L = 0.0188 mol.

    Rate this question:

Zohra Sattar Waxali |PhD (Chemistry) |
Chemistry
Dr. Zohra Sattar Waxali earned her doctorate in chemistry and biochemistry from Northwestern University, specializing in the metallomes of cardiac cells and stem cells, and their impact on biological function. Her research encompasses the development of arsenoplatin chemotherapeutics, stapled peptide estrogen receptor inhibitors, and antimicrobial natural products.

Quiz Review Timeline (Updated): Feb 12, 2024 +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Feb 12, 2024
    Quiz Edited by
    ProProfs Editorial Team

    Expert Reviewed by
    Zohra Sattar Waxali
  • Mar 06, 2016
    Quiz Created by
    Eddy8829
Back to Top Back to top
Advertisement
×

Wait!
Here's an interesting quiz for you.

We have other quizzes matching your interest.