Kaplan MCAT General Chemistry Review

Example: If enough water is added to 11 g of CaCl₂ to make 100 mL of solution, what is the molarity of the solution?

Solution:

Molality

The molality (m) of a solution is the number of moles of solute per kilogram of solvent. For dilute aqueous solutions at 25°C, the molality is approximately equal to molarity, because the density of water at this temperature is 1 kilogram per liter (1 kg/L). However, note that this is an approximation and true only for dilute aqueous solutions. (As aqueous solutions become more concentrated with solute, their densities become significantly different from that of pure water; most water-soluble solutes have molecular weights significantly greater than that of water, so the density of the solution increases as the concentration increases.) You won’t use molality very often, so be mindful of the special situations when it is required: boiling point elevation and freezing point depression.

Example: If 10 g of NaOH are dissolved in 500 g of water, what is the molality of the solution?

Solution:

Normality

We discussed the related concepts of gram equivalent weight, equivalents, and normality (N) in Chapter 4. We would urge you to review those relevant sections. The normality of a solution is equal to the number of equivalents of solute per liter of solution. An equivalent, or gram equivalent weight, is a measure of the reactive capacity of a molecule. Most simply, an equivalent is equal to a mole of charge.

To calculate the normality of a solution, you need to know for what purpose the solution is being used, because it is the concentration of the reactive species with which we are concerned. For example, in acid-base reactions, we are most concerned with the concentration of hydrogen ions; in oxidation-reduction reactions, we are most concerned with the concentration of electrons. Normality is unique among concentration units in that it is reaction-dependent. For example, in acidic solution, 1 mole of the permanganate ion (MnO₄^-) will readily accept 5 moles of electrons, so a 1 M solution would be 5 N. However, in alkaline solution, 1 mole of permanganate will accept only 3 moles of electrons, so in alkaline solution, a 1 M permanganate solution would be 3 N.

MCAT Expertise

Simple ideas on Test Day will make things easier. So, when you come across normality you can think of it as “molarity of the stuff of interest” in the reaction.

DILUTION

A solution is diluted when solvent is added to a solution of high concentration to produce a solution of lower concentration. The concentration of a solution after dilution can be conveniently determined using the equation:

M_iV_i = M_fV_f

where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values, respectively.

MCAT Expertise

This equation is worthy of memorization. Note that it works for any units of concentration, not just molarity, if we replace the M with C for concentration.

Example: How many mL of a 5.5 M NaOH solution must be used to prepare 300 mL of a 1.2 M NaOH solution?

Solution:

Solution Equilibria