Concentration and molarity: calculate molarity, parts per million and percent by mass using the concentration formulas on Table T.

What this topic is asking

The Core Curriculum asks you to express the concentration of a solution in three ways: molarity, parts per million (ppm) and percent by mass. The Regents provides all three formulas on Table T, so this topic is about choosing the right formula and substituting correctly. It builds on the mole and connects forward to titration.

Molarity

Molarity is the most-used concentration unit on the Regents because it connects directly to the mole. If a problem gives a mass of solute, first convert it to moles (using the gram-formula mass) before dividing by the volume in liters. The volume is the volume of the final solution, not the volume of water added.

Parts per million and percent by mass

Parts per million is convenient when a percent would be an awkwardly small number: $8$ ppm is much easier to read than $0.0008\%$. Both formulas are on Table T, and both use grams of solution (solute plus solvent) in the denominator.

Choosing the right measure

Use molarity when the question involves moles or reactions (it links straight to stoichiometry and titration). Use percent by mass for everyday mixtures described by mass. Use parts per million for very dilute solutions. The Regents tells you which to calculate, and Table T supplies the matching formula.

A note on dilution

Adding more solvent to a solution dilutes it: the number of moles of solute stays the same, but the volume increases, so the molarity decreases. Some Regents questions use this idea qualitatively (adding water lowers the concentration) and some use the relationship that moles before equals moles after dilution. The amount of solute does not change just because more water is added, so a more dilute solution simply spreads the same moles through a larger volume.

Why molarity links to reactions

Molarity is the bridge between a measured volume of solution and a number of moles, which is exactly what reaction calculations need. If you know the molarity and the volume of a solution, you can find the moles of solute it contains by multiplying: $\text{moles} = \text{molarity} \times \text{liters}$. This rearrangement of the Table T formula is the step that makes titration work, because it lets you convert a volume you can measure with a buret into the moles you need for a mole-ratio calculation. Concentration is therefore not just a label on a bottle; it is the quantitative entry point to solution chemistry.

Try this

Q1. Calculate the molarity of a solution with $2.0$ mol of solute in $4.0$ L of solution. [1 point]

• Cue. $\dfrac{2.0}{4.0} = 0.50$ M.

Q2. State the parts per million of a solution with $0.0050$ g of solute in $1000.$ g of solution. [1 point]

• Cue. $\dfrac{0.0050}{1000.} \times 1\,000\,000 = 5.0$ ppm.