Molarity and solution stoichiometry: calculate molarity, prepare and dilute solutions, and use molarity in solution stoichiometry.

What this topic is asking

Standard CH.5 asks you to measure and use the concentration of a solution. Virginia expects you to calculate molarity (moles per liter), to prepare and dilute solutions with $M_1 V_1 = M_2 V_2$, and to use molarity in solution stoichiometry to find moles from a volume of solution. Molarity is the bridge between a measured volume of solution and the moles needed for reaction calculations.

Molarity

Molarity is the most common concentration unit in chemistry because it links directly to moles, the unit reactions use. To find the molarity, you need the moles of solute (often from a mass via the molar mass) and the volume of the whole solution in liters. Note that the volume is the volume of the solution, not of the solvent added.

Using molarity to find moles

Rearranging the definition gives the most useful working form:

So a measured volume of a solution of known concentration tells you the moles present. For example, $0.50$ L of a $2.0$ M solution contains $0.50 \times 2.0 = 1.0$ mol of solute. If a volume is given in milliliters, divide by $1000$ to convert to liters first.

Dilution

The dilution equation works because $M \times V$ equals moles, and the moles are conserved. To dilute a stock solution to a target, solve for the volume of stock needed ($V_1$), measure it out, and add solvent up to the final volume ($V_2$). The same equation, in reverse, finds the new concentration when a fixed volume of stock is diluted to a known final volume. Because both volumes appear in the same equation, they only need to share the same unit (both in liters or both in milliliters) rather than being converted to liters, which can save a step.

To prepare a solution of a chosen molarity from a solid, the order of steps matters: calculate the moles needed ($M \times V$), convert moles to a mass with the molar mass, weigh out that mass, dissolve it in less than the final volume, then add solvent up to the mark so the total solution volume is exactly right. Adding water to a fixed mass until the volume is reached, rather than adding water to a fixed volume, is what keeps the molarity accurate.

Try this

Q1. Calculate the molarity of a solution with $3.0$ mol of solute in $1.5$ L of solution. [1 point]

• Cue. $M = \dfrac{3.0}{1.5} = 2.0$ M.

Q2. How many moles of solute are in $0.50$ L of a $4.0$ M solution? [1 point]

• Cue. $\text{moles} = 4.0 \times 0.50 = 2.0$ mol.