Solute and solvent, molarity, dilution, and the percent and parts-per units. The in-class questions, practice and rounds of ten.
Friday 2 October
Solutions, from the Chapter 3 notes: solute and solvent, molarity, dilution, then mass percent, ppm and ppb.
Homework: Molarity and Other Units for Solution Concentrations, in the Chapter 3 folder.
A solution is a homogeneous mixture, the same all the way through, like salt water, air or brass. The solute is what dissolves, the solvent is what it dissolves in, and a concentration says how much solute sits in a given amount of solution.
The component present in the greatest amount. Everything else is dissolved in it.
A component dissolved in the solvent, present in the lesser amount. A solution can have more than one.
Water is the solvent. NaCl(aq) is sodium chloride dissolved in water.
Relative words. A dilute solution holds a relatively small amount of solute, a concentrated one a relatively large amount.
M = moles of soluteliters of solution
5.85 g of sodium chloride, NaCl, is dissolved in water to make 250.0 mL of solution. What is the molarity?
22.99 + 35.45 = 58.44 g/mol
Grams to moles first: divide by the molar mass.
5.85 g ÷ 58.44 g/mol = 0.100 mol
Milliliters to liters, then moles over liters.
250.0 mL = 0.2500 L, 0.100 mol ÷ 0.2500 L = 0.400 M
5.85 g has three significant figures, the fewest in the problem, so the molarity keeps three.
The volume is of the whole solution once it is made up, not of the water that went in.
500 mL of one solution and 500 mL of another do not have to make 1000 mL. Masses always add; volumes need not.
Molarity times liters gives moles, and moles times the molar mass gives the grams to weigh out.
M1V1 = M2V2
What volume of 12.0 M hydrochloric acid is needed to make 500. mL of 0.600 M acid?
Adding water leaves the moles of HCl where they were, so the moles before equal the moles after. Solve for V1.
V1 = M2V2M1 = 0.600 M × 500. mL12.0 M = 25.0 mL
Measure 25.0 mL of the 12.0 M acid and add water up to 500. mL. Both volumes are in milliliters; they only have to be in the same unit.
Diluting adds solvent. Nothing reacts, so dilution is a physical process.
The same number of moles of solute is there before and after, spread through a larger volume, which is why the molarity drops.
12.5 g of sugar is dissolved in 237.5 g of water. The solution weighs both together.
12.5 g + 237.5 g = 250.0 g, 12.5 g ÷ 250.0 g × 100% = 5.00%
A 2.00 kg sample of water holds 1.2 mg of lead. Put both masses in grams before dividing.
1.2 mg = 0.0012 g, 2.00 kg = 2000 g, 0.0012 g ÷ 2000 g × 106 = 0.60 ppm
The mass of a solution is the mass of the solute plus the mass of the solvent. Parts per million and parts per billion suit very small amounts, such as a contaminant in water.
From a unit based on moles to one based on grams, you need the molar mass.
From a unit based on mass to one based on volume, you need the density of the solution.
A sodium chloride solution is 10.0% NaCl by mass and has a density of 1.07 g/mL. What is its molarity? Take 100.0 g of the solution, as for an empirical formula.
100.0 g of solution holds 10.0 g of NaCl. The molar mass turns it into moles.
10.0 g ÷ 58.44 g/mol = 0.171 mol
The density turns the 100.0 g of solution into a volume.
100.0 g ÷ 1.07 g/mL = 93.5 mL = 0.0935 L
Moles over liters, to the three significant figures of 10.0% and 1.07 g/mL.
0.171 mol ÷ 0.0935 L = 1.83 M
The six questions from the Chapter 3 Solutions in-class sheet. Answers are checked for significant figures as well as value, and the working stays hidden until you have tried.
A fresh problem every time, checked box by box for value and significant figures.
Start another round for a fresh set of ten.