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The mass to weigh, the flask to weigh it into, and why the water goes in last — with grams per liter as the same solution's other label.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to work out the mass of a solid needed for a solution of stated concentration and volume, the volume a given mass should be made up to, and the same solution's concentration in grams per liter. You will be able to say why the solid is dissolved before the flask is filled to the mark rather than after, and why three flasks at the same concentration hold the same number of moles and three different masses.
You can work out an amount in moles from a concentration and a volume, a mass from an amount and a molar mass, and a new concentration after a dilution. This lesson adds no relationship to those three. What it adds is the apparatus they are used with, and the two places where a correct calculation still produces the wrong solution.
| Term | What it means |
|---|---|
| Standard solution | A solution whose concentration is known accurately, made from a weighed solid. |
| Volumetric flask | A flask with one mark on its neck that measures one stated volume. |
| Primary standard | A solid pure and stable enough to be weighed and trusted directly. |
| Making up to the mark | Adding water until the liquid's curved surface sits on the line. |
| Stock solution | A concentrated solution kept for diluting as needed. |
To make a standard solution of stated concentration and volume:
$$n = c \times V \qquad\text{then}\qquad m = n \times M$$
with $V$ in liters. Two multiplications, both of which you can already do. What the lesson is really about is the order of the three physical steps that follow, because getting them in the wrong order spoils a solution that the arithmetic says is right.
Step 3 is the one worth saying out loud. A concentration is moles per liter of solution, not per liter of water added. Dissolving $10.6$ g of sodium carbonate in $1$ L of water gives a solution of a little more than a liter, and so a concentration a little below $0.1$ mol/L. The flask exists to fix the volume of the finished solution, and it can only do that if the water goes in last.
Another way: picture
Think of the flask as the answer to the question rather than a container for it. You are not making a liter of water with something in it; you are making a liter of solution, and the solid is part of that liter. The mark is where the finished thing comes up to, which is why nothing may be added after it.
Another way: steps
To make 250 mL of 0.1 mol/L sodium carbonate:
Convert the volume to liters. A volumetric flask is marked in milliliters; divide by 1000.
Find the amount. Concentration times volume in liters.
Find the mass. Amount times the molar mass of the solid you are weighing — the hydrate's molar mass if the bottle holds a hydrate.
Weigh, dissolve, transfer, fill. Weigh into a beaker, dissolve in a little water, pour into the flask, rinse the beaker into the flask two or three times, then add water to the mark and invert to mix.
If the mass is already weighed, run it backwards: mass over molar mass for the moles, moles over concentration for the volume.
For a dilution from stock, find the moles the new flask needs, then the volume of stock that holds them.
Check the work. Is the mass a sensible amount to weigh — between a few tenths of a gram and a few tens of grams for ordinary flasks? Is a calculated volume one that a real flask has: 100, 250, 500 or 1000 mL? Is a stock volume smaller than the flask it goes into? Any answer that fails one of those has a units slip or an inverted division in it.
Multiplying concentration by volume is allowed because concentration is moles per liter; times liters it gives moles.
Multiplying by the molar mass is allowed because every mole of the solid weighs the molar mass in grams, so the moles needed have a definite mass.
Dissolving before filling is required because the concentration is defined per liter of finished solution. The volumetric flask fixes that volume only if the solid is already inside it when the liquid reaches the mark.
Rinsing the beaker into the flask is required because the mass weighed is the mass the calculation assumed. Any solid left behind is solute that never reached the solution.
Taking a smaller volume of stock is allowed because the water added afterward brings no solute with it. The moles in the flask are the moles that came from the stock, so the stock volume is the moles needed divided by the stock's concentration.
A bottle may be labeled in moles per liter or in grams per liter, and the two are the same solution seen by a chemist and by a technician.
$$\text{grams per liter} = \text{moles per liter} \times M$$
because one liter holds $c$ moles and each of them weighs $M$ grams. Going the other way, divide.
| Solution | mol/L | $M$ | g/L |
|---|---|---|---|
| sodium carbonate | $0.1$ | $106$ | $10.6$ |
| sodium chloride | $0.2$ | $58.5$ | $11.7$ |
| potassium hydroxide | $0.5$ | $56$ | $28$ |
Neither number says anything about how much solution there is. A $250$ mL bottle and a $2$ L bottle of the same solution have the same concentration in both units, and different amounts of solute in them.
The reason a calculation wants mol/L is that a balanced equation counts moles. The reason a label often carries g/L is that a balance measures grams. Being able to move between them in one multiplication is most of what makes a stockroom usable.
Not every solid makes a good standard solution, and the reason is worth knowing because it explains a step that otherwise looks like superstition.
A primary standard has to be pure, of known formula, stable in air, and reasonably heavy per mole so that a small weighing error matters little. Sodium carbonate qualifies: it can be dried in an oven and then weighed, and it stays what it was while sitting on the balance.
Sodium hydroxide does not. It absorbs water from the air, and it absorbs carbon dioxide too, so a bottle of it is part sodium hydroxide, part water, part sodium carbonate — in proportions nobody knows. You can weigh out $5$ g of it and have no idea how much sodium hydroxide you have weighed.
So a laboratory makes sodium hydroxide solution up to approximately the concentration it wants, and then finds the real concentration by titrating it against a standard solution of something that could be trusted. That is why the next lesson exists, and why the first sentence of it is that a titration measures a concentration you cannot weigh.
This course still asks you to calculate the mass of sodium hydroxide for a stated concentration, because the calculation is the same calculation. What the calculation cannot tell you is whether the bottle contained what the label said.
Every public water system in the United States must test its water regularly under the Safe Drinking Water Act, and every one of those tests is measured against a standard solution. A lab measuring nitrate, for instance, does not trust its instrument's raw readings. It first runs a set of standards of known concentration, made up exactly as in this lesson, and uses them to calibrate the instrument.
Suppose the lab needs a 1000 mg/L nitrogen standard from potassium nitrate, $\mathrm{KNO_3}$, molar mass 101. Each mole holds 14 g of nitrogen, so 1 g of nitrogen per liter needs $1 \div 14 = 0.0714$ mol per liter. For a 1 L flask that is $0.0714 \times 101 = 7.22$ g of potassium nitrate, dried first so it holds no water. The analyst weighs it, dissolves it, transfers it with rinsings, and fills to the mark.
From that one stock the lab dilutes a series of working standards: 1, 5 and 10 mg/L, each a small measured volume of stock made up to the mark in its own flask. The federal limit for nitrate in drinking water is 10 mg/L as nitrogen, so the standards bracket the number that matters.
Every step is this lesson's arithmetic, and every mistake it warns about would show up as a wrong verdict on a town's water. A solid that had absorbed moisture, or water added to a liter instead of to the mark, would shift every reading the lab reported that day.
The reagent bottles in a home aquarium test kit are standard solutions made up at a factory. The color chart only works because the reagent's concentration is known; a bottle left open long enough to change would give a reading that looks precise and is not.
Water added to the stated volume instead of up to the mark. A liter of water plus a dissolved solid is more than a liter of solution, so the concentration comes out below the one asked for. Dissolve first, fill last.
Concentration treated as a mass per liter. Mass is not amount. Two grams of hydrogen and two grams of lead are the same mass and nothing like the same number of atoms, and a reaction counts atoms. Every question in this unit that looks like it is about mass is about moles with mass at each end. Two flasks at the same concentration hold the same number of moles and, unless the solute is the same substance, different masses.
The flask's size used in a grams-per-liter calculation. It is not in it. Grams per liter is the concentration times the molar mass, whatever size flask the solution happens to be standing in.
Milliliters used as liters. A volumetric flask is marked in milliliters and every formula here wants liters, so there is a division by a thousand at the start of most of these calculations. An answer out by exactly a thousand comes from nowhere else.
A weighed mass assumed to be what the label says. A solid that has taken up water from the air weighs more than the substance in it does, and no calculation downstream can detect it.
Convert to liters.
$500 \div 1000 = 0.5 \text{ L}$
Before multiplying.
Find the amount needed.
$0.25 \times 0.5 = 0.125 \text{ mol}$
Concentration times volume.
Find the molar mass.
$39 + 16 + 1 = 56 \text{ g/mol}$
Potassium hydroxide.
Find the mass to weigh.
$0.125 \times 56 = 7 \text{ g}$
Dissolve, transfer, fill to the mark.
Write the second label.
$0.25 \times 56 = 14 \text{ g/L}$
One more multiplication.
Read the weighed mass.
$14.2 \text{ g of } \mathrm{Na_2SO_4}$
Already on the balance.
Find the molar mass.
$46 + 32 + 64 = 142 \text{ g/mol}$
Sodium sulfate.
Convert to moles.
$14.2 \div 142 = 0.1 \text{ mol}$
Mass over molar mass.
Read the target strength.
$0.05 \text{ mol/L}$
What the solution must be.
Divide for the volume.
$0.1 \div 0.05 = 2 \text{ L}$
Amount over concentration.
Check a flask exists.
$\text{a } 2 \text{ L volumetric flask}$
An impossible volume means reweigh.
Read the stock strength.
$2.0 \text{ mol/L HCl}$
Kept in the stockroom.
Read what is needed.
$500 \text{ mL at } 0.10 \text{ mol/L}$
The working solution.
Convert to liters.
$500 \div 1000 = 0.5 \text{ L}$
Before multiplying.
Find the moles needed.
$0.10 \times 0.5 = 0.05 \text{ mol}$
All from the stock.
Find the stock volume.
$0.05 \div 2.0 = 0.025 \text{ L}$
Moles over stock concentration.
Convert to milliliters.
$0.025 \times 1000 = 25 \text{ mL}$
Measure with a pipet.
Check against the flask.
$25 < 500$
Then fill to the mark with water.
Find the amount needed.
$0.4 \times 0.25 = 0.1 \text{ mol}$
Convert the volume before multiplying.
Find the mass to weigh.
Write the second label.
A bottle is labeled potassium hydroxide, $\mathrm{KOH}$, at $0.5$ mol/L. Its molar mass is $56$ g/mol. What is the same solution's concentration in grams per liter?
Answer: g/L
Complete the worked solution: a technician makes two hundred fifty milliliters of sodium carbonate, $\mathrm{Na_2CO_3}$, at one tenth of a mole per liter. Use sodium at twenty-three, carbon at twelve and oxygen at sixteen. Find the moles needed, the molar mass, and the grams to weigh out.
Find the moles needed.
$\text{one tenth} \times \text{a quarter of a liter} =$ n
Concentration times volume in liters.
Find the molar mass.
$\text{two sodium} + \text{one carbon} + \text{three oxygen} =$ w
Each count times its atomic mass.
Find the mass to weigh.
$(\text{moles}) \times (\text{molar mass}) =$ m
Then dissolve and fill to the mark.
A technician weighs out exactly $5.85$ g of sodium chloride, $\mathrm{NaCl}$, molar mass $58.5$ g/mol, and needs the finished solution to be at $0.2$ mol/L. What volume of solution should it be made up to? Answer in milliliters.
Answer: unit: L / cL / m3 / mL
Three solutions of potassium nitrate, $\mathrm{KNO_3}$, molar mass $101$ g/mol, are to be made up in $0.5$ L flasks at three different concentrations. Work out the amount of solute and the mass to weigh out for each.
| concentration, in mol/L | amount of solute, in mol | mass to weigh out, in g | |
|---|---|---|---|
| Flask A | 0.8 | ||
| Flask B | 0.7 | ||
| Flask C | 0.1 |
A hydroponic lettuce grower in California mixes potassium nitrate, $\mathrm{KNO_3}$ (101 g/mol), into a $300$ L tank at $0.02$ mol/L. How many grams must be weighed out?
The answer: a g.
A water-treatment laboratory keeps a stock solution of sodium chloride, $\mathrm{NaCl}$, at $1$ mol/L. An analyst needs $1000$ mL of it at $0.5$ mol/L for the day's work. What volume of the stock should be measured out before the flask is made up to the mark with water? Answer in milliliters.
Answer: unit: L / cL / m3 / mL
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Three flasks are to be made up, each holding $1$ L of solution at a concentration of $0.3$ mol/L. One holds sodium carbonate, one copper(II) sulfate and one potassium chloride. Work out the amount of solute each flask needs and the mass that has to be weighed out for it. The molar masses are given.
| molar mass, in g/mol | amount of solute, in mol | mass to weigh out, in g | |
|---|---|---|---|
| sodium carbonate, $\mathrm{Na_2CO_3}$ | 106 | ||
| copper(II) sulfate, $\mathrm{CuSO_4}$ | 159.5 | ||
| potassium chloride, $\mathrm{KCl}$ | 74.5 |
You can turn a stated concentration and volume into a mass on a balance, and back again, and you can put the same solution on a label in either of the two units a laboratory uses. Say out loud what would be wrong with a solution made by adding a liter of water to a weighed solid. Next: using a solution you have made up to find the concentration of one you cannot weigh.
15. Your turn: what mass of magnesium sulfate, $\mathrm{MgSO_4}$, molar mass $120$ g/mol, is needed for $250$ mL of a $0.4$ mol/L solution, and what is that in grams per liter?, step 2
$0.1 \times 120 = 12 \text{ g}$
Amount times molar mass.
15. Your turn: what mass of magnesium sulfate, $\mathrm{MgSO_4}$, molar mass $120$ g/mol, is needed for $250$ mL of a $0.4$ mol/L solution, and what is that in grams per liter?, step 3
$0.4 \times 120 = 48 \text{ g/L}$
Concentration times molar mass.