Expressing Concentration of Solutions

Why "Dilute" and "Concentrated" Aren't Enough

In Section 1 we ended on a warning: calling a solution "dilute" or "concentrated" is hopelessly vague. A chemist needs a number. This section is your complete toolkit of those numbers — the units of concentration.

Let's break this down. There are several ways to express the composition of a solution, and each one answers the question "how much solute, relative to what?" in a slightly different way. Some compare masses, some compare moles, some compare to volume. Your job in this chapter — and in every numerical that follows — is to pick the right one and convert fluently between them.

Here's the master list we'll build:

• Mass percentage (w/w) and volume percentage (v/v)
• Mass by volume percentage (w/v)
• Parts per million (ppm)
• Mole fraction ($x$)
• Molarity ($M$)
• Molality ($m$)

Key Point: Two of these — mole fraction and molality — are independent of temperature. The rest involve volume or are tied to it, and since volume changes with temperature, they are temperature-dependent. This single fact is the most tested conceptual point in the section.

Percentage and ppm

Mass percentage (w/w) — mass of solute per 100 g of solution:

$$\text{Mass \%} = \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100$$

A "10% glucose solution by mass" means 10 g glucose in 100 g of solution (i.e. 10 g glucose + 90 g water).

Volume percentage (v/v) — used for liquid-liquid solutions:

$$\text{Volume \%} = \frac{\text{Volume of component}}{\text{Total volume of solution}} \times 100$$

A 35% (v/v) ethylene glycol antifreeze means 35 mL glycol per 100 mL solution.

Mass by volume percentage (w/v) — mass of solute per 100 mL of solution. Common in medicine and pharmacy (e.g. normal saline is 0.9% w/v NaCl).

Parts per million (ppm) — for very dilute solutions (pollutants, trace ions):

$$\text{ppm} = \frac{\text{Mass (or moles) of solute}}{\text{Total mass (or moles) of solution}} \times 10^6$$

[JEE Tip] ppm can be expressed mass-to-mass, volume-to-volume or mass-to-volume — always read which one the problem wants. For dilute aqueous solutions, 1 ppm ≈ 1 mg of solute per litre of water (since 1 L water ≈ 10$^6$ mg).

Mole Fraction

Mole fraction ($x$) compares the moles of one component to the total moles of all components. For a binary solution of solute (B) in solvent (A):

$$x_A = \frac{n_A}{n_A + n_B}, \qquad x_B = \frac{n_B}{n_A + n_B}$$

where $n = \dfrac{\text{given mass}}{\text{molar mass}}$.

The defining property — and the reason mole fraction dominates the colligative-property chapters ahead — is that all the mole fractions add up to 1:

$$x_A + x_B = 1 \quad (\text{and in general } \sum x_i = 1)$$

Because it's a ratio of moles to moles, mole fraction has no units and is independent of temperature.

[NEET Important] Mole fraction is the natural language of Raoult's law and vapour-pressure problems (Sections 4–6). Get comfortable computing it now and those sections become much easier.

Molarity and Molality — Don't Mix Them Up

These two sound almost identical and are the single biggest source of careless errors in the chapter. Learn the difference cold.

Molarity (M) — moles of solute per litre of solution:

$$M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}} = \frac{n_B}{V_{\text{solution (L)}}}$$

Units: mol L$^{-1}$ (also written M). Because it depends on the volume of solution, and volume expands when heated, molarity decreases as temperature rises — it is temperature-dependent.

Molality (m) — moles of solute per kilogram of solvent:

$$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{n_B}{W_{A}\,(\text{kg})}$$

Units: mol kg$^{-1}$ (also written m). It uses mass of solvent, and mass never changes with temperature, so molality is temperature-independent.

Key Point — the one-line memory hook:

• MolaRity → "R" for Litre of solution (volume) → temperature-dependent.
• MolaLity → "L" for kiLogram of solvent (mass) → temperature-independent.

[JEE Tip] For dilute aqueous solutions near room temperature, molarity ≈ molality (because 1 L of dilute solution ≈ 1 kg of water). But never assume this in a calculation unless the problem is explicitly dilute and aqueous.

Interconverting the Units (the Density Bridge)

Exam problems love asking you to convert between molarity, molality, mole fraction and mass %. The bridge that connects mass-based units (molality, mass %, mole fraction) to volume-based ones (molarity) is density:

$$\text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density}}$$

A reliable recipe that never fails:

1. Assume a convenient basis. For a "X% by mass" problem, assume 100 g of solution (so solute = X g, solvent = (100 − X) g).
2. Convert masses to moles using molar masses.
3. For molality: divide moles of solute by kg of solvent.
4. For molarity: find the solution volume using density ($V = \text{mass}/d$), then divide moles of solute by litres of solution.
5. For mole fraction: divide moles of each component by total moles.

[NEET Important] The single most common mistake: using the mass (or volume) of the solution where you needed the solvent, or vice versa. Molality uses solvent; molarity uses solution. Underline which one the formula demands before plugging in.

Solved Examples

Example 1: Mass percentage

A solution is prepared by dissolving 22 g of benzene in 122 g of carbon tetrachloride. Calculate the mass percentage of benzene and of $\mathrm{CCl_4}$.

Solution:

1. Total mass of solution $= 22 + 122 = 144$ g.
2. Mass % of benzene $= \dfrac{22}{144}\times 100 = 15.28\%$.
3. Mass % of CCl$_4$ $= 100 - 15.28 = 84.72\%$.

Answer: benzene 15.28%, CCl$_4$ 84.72%.

Example 2: Mole fraction from mass %

Calculate the mole fraction of benzene ($\mathrm{C_6H_6}$, M = 78 g mol$^{-1}$) in a solution containing 30% benzene by mass in carbon tetrachloride ($\mathrm{CCl_4}$, M = 154 g mol$^{-1}$).

Solution:

1. Basis: take 100 g of solution → benzene = 30 g, CCl$_4$ = 70 g.
2. Moles: $n_{\text{benzene}} = \dfrac{30}{78}=0.385$ mol; $n_{\mathrm{CCl_4}} = \dfrac{70}{154}=0.455$ mol.
3. Mole fraction: $x_{\text{benzene}} = \dfrac{0.385}{0.385+0.455}=\dfrac{0.385}{0.840}=0.458$.

Answer: $x_{\text{benzene}} \approx 0.46$.

Example 3: Molarity, straightforward

Calculate the molarity of a solution containing 5 g of NaOH in 450 mL of solution. (M of NaOH = 40 g mol$^{-1}$)

Solution:

1. Moles of NaOH $= \dfrac{5}{40}=0.125$ mol.
2. Volume $= 450\text{ mL} = 0.450$ L.
3. Molarity $= \dfrac{0.125}{0.450}=0.278$ mol L$^{-1}$.

Answer: 0.278 M.

Example 4: Molarity by dilution

30 mL of 0.5 M $\mathrm{H_2SO_4}$ is diluted to 500 mL. Find the new molarity.

Solution:

1. Use $M_1 V_1 = M_2 V_2$ (moles conserved on dilution).
2. $0.5 \times 30 = M_2 \times 500$.
3. $M_2 = \dfrac{15}{500}=0.03$ mol L$^{-1}$.

Answer: 0.03 M.

Takeaway: On dilution the number of moles of solute is unchanged — only the volume grows. $M_1V_1 = M_2V_2$ is your fastest tool.

Example 5: Molality from mass of solute and solvent

Calculate the molality of a solution containing 20 g of NaOH in 500 g of water. (M of NaOH = 40 g mol$^{-1}$)

Solution:

1. Moles of NaOH $= \dfrac{20}{40}=0.5$ mol.
2. Mass of solvent $= 500\text{ g} = 0.5$ kg.
3. Molality $= \dfrac{0.5}{0.5}=1.0$ mol kg$^{-1}$.

Answer: 1.0 m.

Example 6: Mass of solute for a target molality

Calculate the mass of urea ($\mathrm{NH_2CONH_2}$, M = 60 g mol$^{-1}$) required to make 2.5 kg of a 0.25 molal aqueous solution.

Solution:

1. Molality $0.25 = \dfrac{n_{\text{urea}}}{\text{kg of water}}$. Since no density is given, interpret this as a dilute aqueous solution and take 2.5 kg as the mass of solvent.
2. Moles of urea $= 0.25 \times 2.5 = 0.625$ mol.
3. Mass of urea $= 0.625 \times 60 = 37.5$ g.

Answer: 37.5 g of urea.

Example 7: The full interconversion — molality, molarity and mole fraction together

The density of a 20% (by mass) aqueous KI solution is 1.202 g mL$^{-1}$. Calculate (a) molality, (b) molarity and (c) mole fraction of KI. (M of KI = 166 g mol$^{-1}$, M of H$_2$O = 18 g mol$^{-1}$)

Solution: Basis: 100 g of solution → KI = 20 g, water = 80 g.

• Moles of KI $= \dfrac{20}{166}=0.1205$ mol.
• Moles of water $= \dfrac{80}{18}=4.444$ mol.

(a) Molality $= \dfrac{0.1205}{0.080\text{ kg}} = 1.51$ m.

(b) Molarity: volume of solution $= \dfrac{100\text{ g}}{1.202\text{ g mL}^{-1}}=83.2$ mL $= 0.0832$ L. $$M = \dfrac{0.1205}{0.0832} = 1.45 \text{ mol L}^{-1}.$$

(c) Mole fraction of KI $= \dfrac{0.1205}{0.1205 + 4.444}=\dfrac{0.1205}{4.565}=0.0264$.

Answer: molality ≈ 1.51 m, molarity ≈ 1.45 M, $x_{\text{KI}}\approx 0.026$.

[JEE Tip] Notice molarity (1.45) < molality (1.51) here. Because density > 1 and KI is heavy, the solution volume packs in fewer moles per litre than per kg of solvent. Always sanity-check the relative sizes.

Example 8: ppm of a trace pollutant

A sample of drinking water was found to be severely contaminated with chloroform ($\mathrm{CHCl_3}$). The level of contamination was 15 ppm (by mass). Express this in mass percentage.

Solution:

1. 15 ppm means 15 parts of CHCl$_3$ per $10^6$ parts of solution by mass.
2. Mass % $= \dfrac{15}{10^6}\times 100 = 1.5 \times 10^{-3}\,\%$.

Answer: $1.5\times10^{-3}$ % by mass.

Takeaway: ppm and mass % differ by a factor of $10^4$ ($10^6$ vs $100$). Dividing ppm by $10^4$ gives mass %.

Example 9: Mole fraction must sum to 1

In a binary solution the mole fraction of the solute is 0.2. What is the mole fraction of the solvent?

Solution: Since $x_{\text{solute}} + x_{\text{solvent}} = 1$, we get $x_{\text{solvent}} = 1 - 0.2 = 0.8$.

Answer: 0.8. (A one-line problem — but examiners use it to check you know mole fractions are normalised to 1.)

Example 10: Molarity from mass percent and density

Concentrated nitric acid used in the laboratory is 68% HNO$_3$ by mass in aqueous solution, with density 1.504 g mL$^{-1}$. Calculate its molarity. (M of HNO$_3$ = 63 g mol$^{-1}$)

Solution: Basis: 100 g of solution → HNO$_3$ = 68 g, solution mass = 100 g.

1. Moles of HNO$_3$ $= \dfrac{68}{63}=1.079$ mol.
2. Volume of solution $= \dfrac{100}{1.504}=66.5$ mL $= 0.0665$ L.
3. Molarity $= \dfrac{1.079}{0.0665}=16.23$ mol L$^{-1}$.

Answer: ≈ 16.2 M.

Example 11: Which is temperature-independent?

A student prepares a solution at 25 °C and then warms it to 60 °C. State, with reason, which of its molarity and molality changes.

Solution:

• Molarity changes (decreases) — it is defined per litre of solution, and the solution's volume expands on heating, so the same moles now occupy more litres.
• Molality stays constant — it is defined per kilogram of solvent, and mass is unaffected by temperature.

Answer: molarity decreases; molality is unchanged.

Example 12: Mixing two solutions of the same solute

300 g of a 25% (by mass) solution is mixed with 400 g of a 40% (by mass) solution of the same solute. Calculate the mass percentage of the solute in the final mixture.

Solution:

1. Solute from first solution $= 25\% \text{ of } 300 = 75$ g.
2. Solute from second solution $= 40\% \text{ of } 400 = 160$ g.
3. Total solute $= 75 + 160 = 235$ g.
4. Total mass of mixture $= 300 + 400 = 700$ g.
5. Mass % $= \dfrac{235}{700}\times 100 = 33.57\%$.

Answer: ≈ 33.57% by mass.

Takeaway: When mixing solutions of the same solute, add up grams of solute and grams of total solution separately, then take the ratio. Never average the percentages directly.