To solve molarity practice problems, divide moles of solute by liters of final solution. If the question gives grams, convert them to moles first. If it asks for grams, work backward from concentration and volume. The hardest part is often deciding which quantity the question actually gives you.
This set has eight original problems with worked answers. You’ll practice direct concentration calculations, missing amounts, dilution, and a question that doesn’t contain enough information. These are paper exercises for introductory chemistry. All stated solution volumes refer to the final solution unless a problem explicitly says otherwise.
The three quantities to identify before calculating
Write M = n/V at the top of your page. Here, M is molarity in mol/L, n is the amount of solute in moles, and V is solution volume in liters. Rearranging gives n = MV and V = n/M. The definition and these relationships are explained in OpenStax’s molarity section.
For each question, make three entries: concentration, amount, and final volume. Put a question mark beside the missing value. If the amount is a mass, add a separate conversion line using the supplied molar mass. This small setup keeps a number such as 250 mL from entering your equation as 250 L.
Use 1 L = 1,000 mL, an exact conversion. Keep extra calculator digits until your final answer. The numerical examples below use the significant figures shown in the givens. When a supplied molar mass has more digits than the measurements, it doesn’t determine the final precision.
Try these eight molarity practice problems
Cover the worked answers while you attempt each question. Beside your result, write a sentence explaining whether you multiplied or divided by volume. That sentence is a useful way to catch an equation chosen from memory without checking its meaning.
- A solution contains 0.180 mol of glucose in a final volume of 600. mL. What is its molarity?
- A sample contains 7.31 g of NaCl in 250. mL of solution. Use 58.44 g/mol for NaCl. Find its molarity.
- How many moles of solute are in 35.0 mL of a 0.240 M solution?
- What mass of glucose is present in 150. mL of a 0.200 M glucose solution? Use 180.16 g/mol.
- What solution volume contains 0.0450 mol of solute at a concentration of 0.300 M?
- A 20.0 mL portion of 0.750 M stock solution is diluted to a final volume of 125 mL. Find the final concentration.
- What volume of 1.20 M stock contains the solute needed for 300. mL of 0.0800 M solution?
- A student dissolves 0.100 mol of a substance in 200. mL of water. Can you determine the exact molarity from those facts alone?
Worked answers and the mistake each problem tests
1. Moles and milliliters to molarity
First convert the final volume: 600. mL = 0.600 L. Then calculate M = 0.180 mol / 0.600 L = 0.300 M.
The decimal point in 600. indicates three significant figures for this exercise. Check the meaning: one liter would contain 0.300 mol, so a smaller 0.600 L portion should contain less than that. Multiplying 0.300 mol/L by 0.600 L returns the original 0.180 mol.
A result of 0.000300 M usually means milliliters were used as though they were liters. A result of 0.108 with units mol times L means you multiplied the two givens instead of dividing.
2. Grams to moles before concentration
Convert the mass: n = 7.31 g / 58.44 g/mol = 0.1250856 mol before rounding. Convert the volume: 250. mL = 0.250 L. Then M = 0.1250856 mol / 0.250 L = 0.500 M to three significant figures.
The unit cancellation is the point of the first step: grams divided by grams per mole leaves moles. Dividing 7.31 g directly by 0.250 L produces 29.24 g/L, a mass concentration. That is a different quantity, even though it also describes the solution.
Try a second check using the mass concentration you just found: 29.24 g/L divided by 58.44 g/mol gives the same molarity. Both routes work because the units account for the molar mass. Choose one route for your written answer and use the other to check it.
3. Find the amount in a small portion
Convert 35.0 mL to 0.0350 L. Use n = MV: 0.240 mol/L times 0.0350 L = 0.00840 mol.
This question asks for an amount, not another concentration. The liters cancel during multiplication. The final amount is small because the portion is only 35.0 mL. You can also report it as 8.40 mmol, since one millimole is one thousandth of a mole.
If your result is 6.86, look at its units before accepting it. Dividing concentration by volume produces mol/L squared, which cannot answer a question asking for moles.
4. Concentration to grams
Find the moles first: 0.200 mol/L times 0.150 L = 0.0300 mol. Then convert to mass: 0.0300 mol times 180.16 g/mol = 5.4048 g before rounding. The answer is 5.40 g of glucose.
Keep the identity of the solute next to the mass. A different substance can have the same molarity and volume but a different mass because its molar mass differs. The 180.16 value belongs to glucose; it is not a universal conversion between solution concentration and grams.
For a reverse check, divide 5.4048 g by 180.16 g/mol and then by 0.150 L. You should return to 0.200 mol/L. Use the unrounded mass in that check to avoid introducing a rounding difference.
5. Find the required final volume
Rearrange to V = n/M. Then V = 0.0450 mol / 0.300 mol/L = 0.150 L, or 150. mL.
This is the volume of the solution containing the stated amount at the stated concentration. It isn’t a statement that 150. mL of water should be added to a dry solute. In a course lab, follow the instructor’s procedure for reaching a specified final solution volume.
The division also makes intuitive sense: at a higher target concentration, the same amount of solute would occupy a smaller final volume. If your equation predicts the opposite, revisit the rearrangement.
6. Dilution with a known final volume
For dilution with no solute loss or reaction, the amount of that solute stays constant. Thus M1V1 = M2V2. Calculate M2 = (0.750 M times 20.0 mL) / 125 mL = 0.120 M.
Both volumes can stay in milliliters in this ratio because those units cancel. You can verify the same result in liters: the initial portion contains 0.0150 mol, and 0.0150 mol / 0.125 L = 0.120 M.
The final volume is 125 mL, not 145 mL. The words “diluted to” already describe the total. The dilution factor is 125/20.0 = 6.25, so the final concentration is one sixth-and-a-quarter of the original concentration.
7. Work backward to a stock-solution portion
Use V1 = M2V2/M1. The calculation is (0.0800 M times 300. mL) / 1.20 M = 20.0 mL of stock solution.
Check by counting moles: 0.0200 L times 1.20 mol/L = 0.0240 mol. The target contains 0.300 L times 0.0800 mol/L = 0.0240 mol too. The two amounts agree, which is what a dilution setup must preserve.
The question asks for stock volume, not the volume of water to add. Reaching a final volume of 300. mL is the relevant condition. Don’t automatically describe a precise water volume unless the problem provides an assumption about volume additivity.
8. Recognize missing information
No exact molarity follows from the given facts. You know the amount of solute and the initial water volume, but the final solution volume is missing. Dissolving a substance doesn’t guarantee that the final volume equals the starting water volume.
If an exercise explicitly tells you to assume a final solution volume of 200. mL, then 0.100 mol / 0.200 L gives 0.500 M under that assumption. State the assumption beside the answer. Without it, write “final solution volume needed” instead of inventing a value.
Turn mistakes into a useful correction sheet
Create four columns in your own notes: question number, first setup, corrected setup, and reason for the correction. Use specific labels such as “grams were not converted,” “mL used as L,” or “final volume confused with added water.” A page of those distinctions is more useful for your next attempt than eight copied answers.
Next, change one given in a question and predict the direction of the answer before calculating. For problem 3, doubling the portion volume should double the amount while leaving concentration unchanged. For problem 1, doubling the final volume while preserving the solute amount should halve concentration.
Find complementary chemistry notes for your course. If your next topic combines solution amounts with reactions, review balancing chemical equations practice before using reaction coefficients. Molarity converts volume to moles; a balanced equation provides a separate relationship between reacting substances.
Frequently asked questions
Is molarity the same as moles?
No. Moles measure an amount. Molarity measures amount per solution volume. Two bottles can have the same molarity but different total moles because their volumes differ.
Must I always convert milliliters to liters?
Use liters when calculating M = n/V with moles. In M1V1 = M2V2, matching volume units cancel, so both volumes can be in milliliters.
Can M1V1 = M2V2 solve every solution question?
No. It describes equal amounts of the same solute before and after the change. If a reaction consumes that solute or more is added, establish the new amount first.
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