Expressing the concentration of solutions

What is the concentration of a solution?

In plain words: “a little sugar in water” could mean anything. A concentration puts a number on it, and the right number depends on the job: percentages for labels, ppm for traces, mole fraction for vapour pressure, molarity for measuring out by volume, and molality when the temperature changes.

Mass percentage and volume percentage

• Mass percentage (w/w) of a component is its mass as a percentage of the total mass of the solution (NCERT Eq. (1.1), page 2). With w_A the mass of component A and w the total mass of the solution:

• So 10% glucose in water by mass means 10 g of glucose dissolved in 90 g of water, making 100 g of solution. Mass percentage is common in industry: NCERT’s example of a commercial bleaching solution contains 3.62 mass percentage of sodium hypochlorite in water (NCERT page 2).
• Volume percentage (V/V) is the volume of a component as a percentage of the total volume of the solution (NCERT Eq. (1.2), page 2). With V_A the volume of component A and V the total volume of the solution:

• So 10% ethanol in water means 10 mL of ethanol made up with water to 100 mL of solution. Solutions of liquids are commonly given this way: NCERT gives a 35% (v/v) ethylene glycol solution as an antifreeze example in car engines, where it lowers the freezing point of water to 255.4 K (NCERT page 3).
• NCERT’s pair, 255.4 K and −17.6 °C, uses T(°C) = T(K) − 273; with 273.15, 255.4 K is −17.75 °C, about −17.8 °C (reasoned).
• Note: the volume of the solution is not always the sum of the volumes mixed, which is why volume percentage is defined using the final volume of solution (reasoned from NCERT page 3).

Mass by volume percentage

• Mass by volume percentage (w/V) is the mass of solute, in grams, dissolved in 100 mL of the solution. It is commonly used in medicine and pharmacy (NCERT page 3).
• So a 5% (w/V) glucose solution has 5 g of glucose in every 100 mL of solution (reasoned from NCERT page 3).

Parts per million

• For solutes present in trace quantities, concentration is given in parts per million (ppm) (NCERT Eq. (1.3), page 3):

• NCERT lists ppm on mass-to-mass, volume-to-volume or mass-to-volume reporting bases (NCERT page 3).
• A litre of sea water, of mass 1030 g, contains about 6 × 10⁻³ g of dissolved oxygen: 5.8 g per 10⁶ g of sea water, or 5.8 ppm (NCERT page 3; worked in Q4 below). Pollutants in water are often given in μg mL⁻¹ or ppm (NCERT page 3); for dilute water these are nearly equal, since 1 mL of water has a mass of about 1 g (reasoned).

For a dimensionless ppm ratio, numerator and denominator must use the same kind of quantity. A mass-per-volume report such as mg L⁻¹ equals the numerical mass-based ppm only approximately for a dilute aqueous solution of density near 1 kg L⁻¹; state the basis rather than treating them as universally interchangeable.

Mole fraction

• The mole fraction x of a component is its number of moles divided by the total number of moles of all the components; the subscript names the component (NCERT Eq. (1.4), page 3).
• In a binary solution of A and B (NCERT Eq. (1.5), page 3):

• The mole fractions of all the components add up to one (NCERT Eq. (1.7), page 3):

• Mole fraction is very useful for relating physical properties, such as vapour pressure, to concentration, and for gas mixtures (NCERT page 3).

Molarity

• Molarity (M) is the number of moles of solute dissolved in one litre, or one cubic decimetre, of solution (NCERT Eq. (1.8), page 4). With n_B the moles of solute and V the volume of the solution in litres:

• So a 0.25 mol L⁻¹ (0.25 M) solution of NaOH has 0.25 mol of NaOH in one litre of solution (NCERT page 4). The molar mass turns grams into moles.
• Dilution. Adding solvent changes the volume, not the moles of solute, so M₁V₁ = M₂V₂, as in Class 11.
• NCERT slip (page 4): the working of NCERT Example 1.2 says “Using equation (2.8)”. The molarity equation is Eq. (1.8).

Molality

• Molality (m) is the number of moles of solute per kilogram of solvent (NCERT Eq. (1.9), page 5):

• So a 1.00 mol kg⁻¹ (1.00 m) solution of KCl has 1 mol, 74.5 g, of KCl dissolved in 1 kg of water (NCERT page 5).
• Temperature. NCERT says mass %, ppm, mole fraction and molality do not depend on temperature, but molarity does, because volume changes with temperature and mass does not (NCERT page 5). This assumes fixed composition, without reaction or loss by evaporation. Volume-based concentrations generally vary with temperature; volume percentage can also vary if the component and solution expand differently. Always specify its reference temperature.
• Converting (beyond the NCERT text): for a solution of density d (g mL⁻¹), molarity M and molar mass of solute M_B (g mol⁻¹), one litre has a mass of 1000d g, of which M × M_B g is solute, so m = 1000M/(1000d − M × M_B). In practice, start from a convenient amount, such as 100 g of solution, as in Example 4.

Solved examples

In Example 4, the reported KI mole fraction is the analytical composition based on formula units added. Dissolved KI largely dissociates into ions; fractions based on the actual ionic species would count those separately, as needed in colligative-property models.

Common mistakes to avoid

Quick revision summary

• Mass % = mass of component ÷ mass of solution × 100; volume % uses volumes; w/V % is grams of solute per 100 mL of solution.
• ppm = parts of component ÷ total parts × 10⁶, for traces.
• Mole fraction x_A = n_A/(n_A + n_B); all mole fractions add to 1.
• Molarity: moles per litre of solution. Molality: moles per kg of solvent.
• At fixed composition, mass %, mass-based ppm, mole fraction and molality are temperature-independent; volume-based measures generally vary.

Questions students ask

What is the molarity of pure water?

Pure water is about 55.5 M: a litre of it has a mass of about 1000 g, and 1000 g ÷ 18.02 g mol^⁻¹ = 55.5 mol, the figure NCERT uses for a litre of water in Henry’s law sums (NCERT Example 1.4, page 8). More exactly, water’s density at 25 °C is 0.997 g mL^⁻¹ (beyond the NCERT text; NIST Chemistry WebBook), so a litre holds 997/18.02 = 55.3 mol, and the value falls over temperature ranges where water expands; water near 0–4 °C is a familiar exception to simple expansion on warming. Pure water has no solute, so this is its molar concentration rather than the molarity of a solution; a retained 1 kg sample of water contains the same amount of H₂O when temperature changes.

Why is molality preferred over molarity?

Molality is preferred over molarity when the temperature changes, because it is made up from two masses, which do not change with temperature, while molarity depends on the volume of the solution, which does (NCERT page 5). Beyond the NCERT text, water’s density falls from 0.998 g mL^⁻¹ at 20 °C to 0.992 g mL^⁻¹ at 40 °C (NIST Chemistry WebBook), so a dilute aqueous solution expands by about 0.6% and its molarity falls by about 0.6%, while its molality stays the same. The boiling-point and freezing-point formulas use molality too, as a solution made up at room temperature is measured near its boiling or freezing point (reasoned).

How are molality and mole fraction related?

In a binary solution the solute’s molality and mole fraction x₂ are related by m = 1000x₂/(x₁M₁), where x₁ = 1 − x₂ and M₁ is the solvent’s molar mass in g mol^⁻¹ (beyond the NCERT text, from NCERT Eq. (1.5) and Eq. (1.9)). It holds because 1 kg of solvent is 1000/M₁ mol, with x₂/x₁ mol of solute per mole of solvent. So an aqueous solution with x₂ = 0.050 has m = 1000 × 0.050/(0.950 × 18) = 2.92 mol kg^⁻¹; in a dilute solution x₁ ≈ 1, so the molality is nearly proportional to the mole fraction.

What are the SI units of molarity and molality?

The SI unit of molarity is the mole per cubic metre (mol m^⁻³), with 1 M = 1 mol L^⁻¹ = 1 mol dm^⁻³ = 1000 mol m^⁻³, and that of molality is the mole per kilogram (mol kg^⁻¹); NCERT uses mol L^⁻¹ and mol kg^⁻¹ (NCERT pages 4–5). Beyond the NCERT text, the NIST Guide to the SI calls molarity amount-of-substance concentration, c, and calls the name molarity, the term molal and the unit symbols M and m (as in 0.25 M and 1.00 m) obsolete, while keeping molality in mol kg^⁻¹ (NIST Special Publication 811, Sections 8.6.5 and 8.6.8). Mass and volume percentages, ppm by mass or by volume and mole fraction have no unit.

Is a 5% (w/V) solution also 5% by mass?

A 5% (w/V) solution is 5% by mass only if its density is exactly 1 g mL^⁻¹: it has 5 g of solute in 100 mL of solution (NCERT page 3), but 100 mL of solution has a mass of 100d g, where d is its density in g mL^⁻¹. So for a 5% (w/V) glucose solution of density 1.02 g mL^⁻¹, 100 mL has a mass of 102 g, and the mass percentage is 5/102 × 100 = 4.9%. In general, mass % = (w/V %)/d, so the two are close for dilute aqueous solutions and differ more as the density moves away from 1 g mL^⁻¹ (reasoned).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Find the mass percentage of each (NCERT Intext Question 1.1).

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   Total mass = 144 g. Benzene: 22/144 × 100 = 15.28%; carbon tetrachloride: 122/144 × 100 = 84.72% (worked here).

2. Q2. Find the mole fraction of benzene in a solution containing 30% benzene by mass in carbon tetrachloride (NCERT Intext Question 1.2).

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   In 100 g: benzene 30/78 = 0.385 mol; CCl₄ 70/154 = 0.455 mol. x_benzene = 0.385/(0.385 + 0.455) = 0.458 (worked here). NCERT’s answer key (page 30) prints 0.459 and 0.541. Rounding benzene up to 0.385 mol while keeping 0.4545 mol of CCl₄ gives 0.385/(0.385 + 0.4545) = 0.459; unrounded, 0.3846/(0.3846 + 0.4545) = 0.458, the difference is only in rounding, and the given 30% does not justify treating that last digit as a material disagreement.

3. Q3. Find the molarity of (a) 30 g of Co(NO₃)₂·6H₂O in 4.3 L of solution, (b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL (NCERT Intext Question 1.3).

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   (a) Molar mass = 59 + 2 × 62 + 6 × 18 = 291 g mol⁻¹, so 30 g is 0.103 mol; 0.103/4.3 = 0.024 M. (b) Moles = 0.030 × 0.5 = 0.015 mol; 0.015/0.500 = 0.030 M (worked here).