Concentration units — mass %, volume %, ppm, ppb, molarity (M), molality (m), mole fraction: Exercises

Recall The 7 fractions (open only if stuck on which formula)

mass % $= \frac{mass solute}{mass solution} \times 100$
vol % $= \frac{vol solute}{vol solution} \times 100$
ppm $= \frac{mass solute}{mass solution} \times 1 0^{6}$
$M = \frac{n _{solute}}{V _{solution (L)}}$
$m = \frac{n _{solute}}{mass solvent (kg)}$
$x_{A} = \frac{n _{A}}{n _{total}}$
$m = \frac{M \times 1000}{1000 ρ - M M _{w}}$

Level 1 — Recognition

Exercise L1-1

Dissolve 15 g of sugar in 85 g of water. What is the mass percent of sugar?

Recall Solution

WHAT we need: mass % $= \frac{mass solute}{mass solution} \times 100$ .

Mass of solution $= 15 + 85 = 100$ g. Why add? solution = solute + solvent.
$mass % = \frac{15}{100} \times 100 = 15%$ .

Exercise L1-2

0.5 mol of glucose is dissolved to make 2 L of solution. What is the molarity?

Recall Solution

Molarity uses litres of solution in the denominator.
$M = \frac{n}{V} = \frac{0.5}{2} = 0.25 M$ .

Exercise L1-3

Which units are temperature-independent: molarity, molality, volume %, mole fraction?

Recall Solution

Anything built on mass or moles only never changes with temperature. Anything built on volume expands when heated.

Temperature-independent: molality and mole fraction.
Temperature-dependent: molarity and volume % (both use volume).

Level 2 — Application

Exercise L2-1

5.85 g of NaCl ( $M_{w} = 58.5$ ) is dissolved to make 250 mL of solution. Find the molarity.

Recall Solution

Step 1 — moles. $n = \frac{mass}{M _{w}} = \frac{5.85}{58.5} = 0.1$ mol. Why? grams ÷ molar mass = moles. Step 2 — litres. $250 mL = 0.25$ L. Step 3 — molarity. $M = \frac{0.1}{0.25} = 0.4 M$ .

Exercise L2-2

0.2 mol of urea is dissolved in 400 g of water. Find the molality.

Recall Solution

Molality demands kg of solvent: $400 g = 0.4$ kg.
$m = \frac{n}{kg solvent} = \frac{0.2}{0.4} = 0.5 m$ .

Exercise L2-3

A water sample contains 3 mg of lead in 1.5 L of water ( $ρ = 1$ g/mL). Express the concentration in ppm.

Recall Solution

Step 1 — mass of solution. $1.5 L = 1500 mL$ , so mass $= 1500 \times 1 = 1500$ g $= 1.5 \times 1 0^{6}$ mg. Why? density 1 g/mL turns mL into g. Step 2 — ppm (mass/mass). Keep both in mg: $ppm = \frac{3 mg}{1.5 \times 1 0 ^{6} mg} \times 1 0^{6} = 2 ppm .$ Check with the water shortcut: $\frac{3 mg}{1.5 L} = 2$ mg/L $\approx 2$ ppm ✓ (valid only because $ρ \approx 1$ ).

Exercise L2-4

A mixture contains 2 mol of $N_{2}$ and 3 mol of $O_{2}$ . Find the mole fraction of each.

Recall Solution

$n_{total} = 2 + 3 = 5$ mol.
$x_{N_{2}} = \frac{2}{5} = 0.4$ , $x_{O_{2}} = \frac{3}{5} = 0.6$ .
Check: $0.4 + 0.6 = 1$ ✓ (mole fractions always sum to 1).

Level 3 — Analysis

Exercise L3-1

Concentrated nitric acid is 68% by mass $HNO_{3}$ ( $M_{w} = 63$ ) with density $1.4$ g/mL. Find its molarity.

Recall Solution

Pick a basis: 100 g of solution (natural for mass %).

Mass of $HNO_{3} = 68$ g $\Rightarrow n = \frac{68}{63} = 1.079$ mol. Why 100 g? mass % is "per 100 g solution", so 100 g makes the solute mass literally the percentage.
Volume of that 100 g $= \frac{mass}{ρ} = \frac{100}{1.4} = 71.43$ mL $= 0.07143$ L. Why divide by density? mass → volume.
$M = \frac{1.079}{0.07143} = 15.1 M$ .

Exercise L3-2

You have 250 mL of 0.5 M HCl. How much water must you add to dilute it to 0.2 M? (Assume volumes add for a dilute aqueous mix.)

Recall Solution

Tool: dilution law $M_{1} V_{1} = M_{2} V_{2}$ — see Stoichiometry and Dilution. Why it works: the moles of solute don't change when you add water, and moles $= M V$ .

$M_{1} V_{1} = 0.5 \times 250 = 125$ mmol·mL-units.
$V_{2} = \frac{M _{1} V _{1}}{M _{2}} = \frac{125}{0.2} = 625$ mL (final total volume).
Water added $= 625 - 250 = 375 mL$ .

Exercise L3-3

A 10% (by mass) NaOH ( $M_{w} = 40$ ) solution has density $1.1$ g/mL. Find its molality.

Recall Solution

Basis: 100 g of solution (mass % → use 100 g).

Solute NaOH $= 10$ g $\Rightarrow n = \frac{10}{40} = 0.25$ mol.
Solvent (water) $= 100 - 10 = 90$ g $= 0.09$ kg. Why subtract? solvent = solution − solute; molality needs solvent mass, not solution.
$m = \frac{0.25}{0.09} = 2.78 m$ . Notice density was a decoy — molality never needs volume, so $ρ$ isn't used here.

Level 4 — Synthesis

Exercise L4-1

A 2 M $H_{2} SO_{4}$ solution has density $1.14$ g/mL, $M_{w} = 98$ . Find its molality by rebuilding the formula from a 1 L basis (do not just quote it).

Recall Solution

Basis: 1 L (1000 mL) of solution.

Step 1 — moles of solute: $n = M \times 1 L = 2$ mol.
Step 2 — mass of solution: $1000 \times 1.14 = 1140$ g.
Step 3 — mass of solute: $2 \times 98 = 196$ g.
Step 4 — mass of solvent: $1140 - 196 = 944$ g $= 0.944$ kg.
Step 5 — molality: $m = \frac{2}{0.944} = 2.12 m$ . Matches the boxed formula $m = \frac{M \times 1000}{1000 ρ - M M _{w}} = \frac{2000}{944} = 2.12$ ✓.

Exercise L4-2

For the same 2 M $H_{2} SO_{4}$ (from L4-1), find the mole fraction of $H_{2} SO_{4}$ . Take water's molar mass as 18.

Recall Solution

Reuse the 1 L basis pieces: solute $= 2$ mol; solvent water $= 944$ g.

Moles of water $= \frac{944}{18} = 52.44$ mol. Why? grams ÷ molar mass = moles.
$x_{H_{2} SO_{4}} = \frac{2}{2 + 52.44} = \frac{2}{54.44} = 0.0367$ .
Check: $x_{water} = 1 - 0.0367 = 0.9633$ , and $\frac{52.44}{54.44} = 0.9633$ ✓.

Exercise L4-3

A solution is $0.25$ mole fraction of ethanol ( $M_{w} = 46$ ) in water ( $M_{w} = 18$ ). Find its molality.

Recall Solution

Basis: 1 mol of total solution (natural for mole fractions).

Ethanol (solute) $= 0.25$ mol; water (solvent) $= 0.75$ mol.
Mass of water $= 0.75 \times 18 = 13.5$ g $= 0.0135$ kg. Why water is solvent? it is the larger amount / the medium.
$m = \frac{n _{ethanol}}{kg water} = \frac{0.25}{0.0135} = 18.52 m$ .

Level 5 — Mastery

Exercise L5-1

Commercial concentrated $HCl$ is 37% by mass ( $M_{w} = 36.5$ ), density $1.19$ g/mL. Find (a) its molarity, (b) its molality, and (c) explain in one line why $m > M$ .

Recall Solution

Basis: 100 g of solution (mass % friendly).

Solute HCl $= 37$ g $\Rightarrow n = \frac{37}{36.5} = 1.0137$ mol.
(a) Molarity: volume of 100 g $= \frac{100}{1.19} = 84.03$ mL $= 0.08403$ L. $M = \frac{1.0137}{0.08403} = 12.06 M .$
(b) Molality: solvent $= 100 - 37 = 63$ g $= 0.063$ kg. $m = \frac{1.0137}{0.063} = 16.09 m .$
(c) Here $m > M$ because the solvent (0.063 kg) is far less than 1 kg, and dividing the same moles by a smaller denominator gives a larger number; density $> 1$ also shrinks the volume in $M$ but the small-solvent effect dominates.

Exercise L5-2

You must prepare 500 mL of 0.1 M $Na_{2} CO_{3}$ ( $M_{w} = 106$ ) by diluting a 2 M stock. (a) What volume of stock is needed? (b) How many grams of $Na_{2} CO_{3}$ end up in the final flask?

Recall Solution

(a) Dilution law $M_{1} V_{1} = M_{2} V_{2}$ (moles conserved): $V_{1} = \frac{M _{2} V _{2}}{M _{1}} = \frac{0.1 \times 500}{2} = 25 mLofstock .$ (b) Moles in final flask $= M_{2} V_{2} = 0.1 \times 0.5 = 0.05$ mol. $mass = n \times M_{w} = 0.05 \times 106 = 5.3 g .$ Cross-check: the same 0.05 mol came from 25 mL of 2 M stock: $2 \times 0.025 = 0.05$ mol ✓ — dilution adds water, not solute.

Exercise L5-3

Seawater is about 35 g of dissolved salt per litre, density $1.025$ g/mL. Express this as (a) mass % and (b) ppm.

Recall Solution

Basis: 1 L of seawater.

Mass of solution $= 1000 \times 1.025 = 1025$ g. Why? density × volume = mass.
Solute salt $= 35$ g.
(a) mass %: $\frac{35}{1025} \times 100 = 3.41%$ .
(b) ppm: $\frac{35}{1025} \times 1 0^{6} = 34146 ppm$ (i.e. $\approx 3.41 \times 1 0^{4}$ ppm). Note the naive "35 mg... per L" shortcut fails here: the density is $1.025 \neq = 1$ , so ppm ( $34146$ ) is slightly less than the mg/L figure would suggest.