Mass per Volume Concentration

Introduction

Mass per volume concentration compares a named solute's mass with the final liquid volume. It is convenient when a sample volume is measured directly and the solute amount is expressed in grams or milligrams. The denominator is the complete solution volume, not the volume of pure solvent poured at the start.

Core explanation

Mass concentration ρₛ = m(solute)/V(solution) can be reported in g L⁻¹ or mg L⁻¹. If 3.00 g solute is dissolved and the final volume is 0.250 L, ρₛ = 3.00/0.250 = 12.0 g L⁻¹. Taking a well-mixed 50.0 mL portion gives 12.0 g L⁻¹ × 0.0500 L = 0.600 g solute. The volume portion contains a proportional amount because the solution is homogeneous.

Final volume matters. Dissolving a solid in 250 mL water does not guarantee exactly 250 mL final solution; solute and solvent occupy space in ways that may change total volume. A preparation can specify “dissolve and make up to 250 mL,” which means use a volumetric flask or other suitable measurement after dissolution. That instruction is different from adding solute to 250 mL water.

Units provide a quick check. If solute mass is in mg and volume in L, the answer is mg L⁻¹. One g L⁻¹ equals 1000 mg L⁻¹. To compare with molarity, divide g L⁻¹ by molar mass in g mol⁻¹, yielding mol L⁻¹. For example, 58.5 g L⁻¹ NaCl corresponds to about 1.00 mol L⁻¹ in formula amount when its molar mass is 58.5 g mol⁻¹. That does not automatically state each ion's activity.

Mass per volume is not mass percent. Percent by mass uses solution mass, whereas g L⁻¹ uses solution volume. A density is needed to connect them if the same sample's total mass and volume are not both known. The common shorthand “5% w/v” in some contexts means 5 g per 100 mL final solution, but one must read the stated convention; it is not 5% by mass.

If solvent evaporates while solute remains, final volume decreases and mass concentration increases, until solubility or another process changes the dissolved amount. If a chemical reaction consumes the named solute, use its remaining mass rather than initial mass. In dilute environmental samples, mg L⁻¹ is common, but one mg L⁻¹ is only approximately one part per million by mass when solution density is close to 1 kg L⁻¹.

Step-by-step reasoning

1. Identify the named solute and its mass in the sampled liquid. 2. Obtain the final solution volume and convert it to litres. 3. Divide mass by that volume and attach g L⁻¹ or mg L⁻¹. 4. To find mass in an aliquot, multiply concentration by aliquot volume. 5. Use molar mass or density explicitly if converting to another concentration basis.

Visual explanation

Draw a volumetric flask with a 0.250 L mark and a label “3.00 g solute in the entire filled flask.” An arrow to a 0.0500 L pipette portion shows one fifth of the solution volume and one fifth of the solute mass.

Real-world analogy

A recipe may specify twelve grams of ingredient per litre of finished drink. Measuring one quarter litre gives one quarter of that ingredient when well mixed. Counting the water poured before other ingredients were added would use the wrong finished-drink volume.

Real-world example

Water analysis may report dissolved nitrate as mg L⁻¹. A laboratory must identify whether it means nitrate ion mass or nitrogen mass in nitrate, since these have different molar masses. A number without the reported species is incomplete for chemical calculations.

Why?

Why does the denominator refer to solution rather than solvent volume? The concentration describes what is in each unit volume of the liquid sample actually collected or delivered, after the solute has been incorporated.

Common misconception

“g L⁻¹ is the same as mol L⁻¹.” Grams and moles are different numerators. Convert with the solute's molar mass before treating mass concentration as molarity.

Worked example

A solution contains 2.50 g CuSO₄ in 500.0 mL final solution. Its mass concentration is 2.50/0.5000 = 5.00 g L⁻¹. A 20.0 mL aliquot contains 5.00 × 0.0200 = 0.100 g CuSO₄ formula mass. If the same 2.50 g had been added to 500.0 mL water, the actual final volume would need measurement before an exact g L⁻¹ value could be assigned.

Quick check

1. What mass of solute is in 0.200 L of a 15.0 g L⁻¹ solution? Answer: Multiply concentration by solution volume: 15.0 × 0.200 = 3.00 g solute.

Exam focus

Convert mL to L and identify the chemical species represented by the mass. Keep g L⁻¹ distinct from molarity and mass percent.

Advanced insight

At low concentration in water, numerical mg L⁻¹ and parts per million by mass may be close because water density is near 1 kg L⁻¹. At other densities or high concentrations the shortcut can be appreciably wrong.

Summary

Mass concentration is solute mass per final solution volume. Its units reveal the numerator and denominator. Representative aliquots scale with volume, while conversions to molarity or mass percent require molar mass or density information.

Practice questions

1. Calculate g L⁻¹ for 1.20 g solute in 300 mL final solution. Answer: 300 mL = 0.300 L, so 1.20/0.300 = 4.00 g L⁻¹. 2. How many milligrams are in 50.0 mL of a 20 mg L⁻¹ solution? Answer: 50.0 mL = 0.0500 L, so mass is 20 × 0.0500 = 1.0 mg. 3. What extra property is needed to turn g L⁻¹ into mol L⁻¹? Answer: The named solute's molar mass, in g mol⁻¹, converts grams into moles.