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Chemistry: Stoichiometry - Mass-by-Volume and Volume-by-Volume Concentration

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Concentration refers to the amount of solute present in a given quantity of solution or solvent.

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Mass-by-Volume Percentage (%m/v\% m/v) expresses the concentration of a solution as the mass of the solute (in grams) dissolved in 100 mL100 \, mL of the solution. It is commonly used in medicine and pharmacy.

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Volume-by-Volume Percentage (%v/v\% v/v) expresses the concentration as the volume of a liquid solute (in mLmL) dissolved in 100 mL100 \, mL of the total solution. This is used when both the solute and solvent are liquids.

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The total volume of a solution is often the sum of the volumes of the solute and the solvent, expressed as: Volume of Solution=Volume of Solute+Volume of Solvent\text{Volume of Solution} = \text{Volume of Solute} + \text{Volume of Solvent}.

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In stoichiometry calculations, always ensure units are consistent (e.g., mass in grams and volume in milliliters) before applying the percentage formulas.

📐Formulae

% Mass-by-Volume (m/v)=(Mass of Solute (g)Volume of Solution (mL))×100\% \text{ Mass-by-Volume (m/v)} = \left( \frac{\text{Mass of Solute (g)}}{\text{Volume of Solution (mL)}} \right) \times 100

% Volume-by-Volume (v/v)=(Volume of Solute (mL)Volume of Solution (mL))×100\% \text{ Volume-by-Volume (v/v)} = \left( \frac{\text{Volume of Solute (mL)}}{\text{Volume of Solution (mL)}} \right) \times 100

Volume of Solution (mL)=Volume of Solute (mL)+Volume of Solvent (mL)\text{Volume of Solution (mL)} = \text{Volume of Solute (mL)} + \text{Volume of Solvent (mL)}

💡Examples

Problem 1:

A saline solution is prepared by dissolving 15 g15 \, g of Sodium Chloride (NaClNaCl) in enough water to make a 300 mL300 \, mL solution. Calculate the mass-by-volume percentage of the solution.

Solution:

Mass of Solute=15 g\text{Mass of Solute} = 15 \, g Volume of Solution=300 mL\text{Volume of Solution} = 300 \, mL Percentage (m/v)=15300×100\text{Percentage (m/v)} = \frac{15}{300} \times 100 Percentage (m/v)=0.05×100=5%\text{Percentage (m/v)} = 0.05 \times 100 = 5\%

Explanation:

To find the mass-by-volume percentage, we divide the mass of the salt (15 g15 \, g) by the total volume of the solution (300 mL300 \, mL) and multiply by 100100.

Problem 2:

If 40 mL40 \, mL of ethanol is mixed with 160 mL160 \, mL of water to form a solution, what is the volume-by-volume percentage of ethanol in the solution?

Solution:

Volume of Solute (ethanol)=40 mL\text{Volume of Solute (ethanol)} = 40 \, mL Volume of Solvent (water)=160 mL\text{Volume of Solvent (water)} = 160 \, mL Total Volume of Solution=40+160=200 mL\text{Total Volume of Solution} = 40 + 160 = 200 \, mL Percentage (v/v)=40200×100\text{Percentage (v/v)} = \frac{40}{200} \times 100 Percentage (v/v)=0.2×100=20%\text{Percentage (v/v)} = 0.2 \times 100 = 20\%

Explanation:

First, calculate the total volume of the solution by adding the volume of solute and solvent. Then, divide the volume of the solute (40 mL40 \, mL) by the total solution volume (200 mL200 \, mL) and multiply by 100100.

Problem 3:

How many grams of glucose are required to prepare 500 mL500 \, mL of a 10% (m/v)10\% \, (m/v) glucose solution?

Solution:

Percentage (m/v)=10%\text{Percentage (m/v)} = 10\% Volume of Solution=500 mL\text{Volume of Solution} = 500 \, mL 10=Mass of Solute500×10010 = \frac{\text{Mass of Solute}}{500} \times 100 10=Mass of Solute510 = \frac{\text{Mass of Solute}}{5} Mass of Solute=10×5=50 g\text{Mass of Solute} = 10 \times 5 = 50 \, g

Explanation:

By rearranging the mass-by-volume formula, we can solve for the mass of the solute. Multiplying the percentage by the volume and dividing by 100100 gives the required mass of glucose.