Ch. 2 | Solutions0/13

Expressing Concentration — Part 1

Mass percentage, volume percentage, and mole fraction

Quantitative ReasoningLevel 2 · Application

A nurse is asked to prepare a 5% w/v glucose solution for a patient drip. She dissolves 5 g of glucose in 100 mL of water. A second nurse dissolves 5 g of glucose in enough water to make 100 mL of final solution. Only one nurse prepared the correct 5% w/v solution. Which one, and why does the difference matter?

Real Life Hook

Breathalysers measure blood alcohol content (BAC) in grams of ethanol per 100 mL of blood — a w/v percentage. In India, the legal driving limit is 0.03% w/v BAC (30 mg per 100 mL). At 0.08%, coordination and reaction time are significantly impaired. The difference between legal and dangerous is a fraction of a percent — which is why standardised concentration units are not just academic formality.

Mass Percentage (w/w %)

The most common way to express concentration of solid-in-liquid or liquid-in-liquid solutions.

w/w%=mass of solutemass of solution×100\text{w/w\%} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

Mass of solution = mass of solute + mass of solvent

Example: 10 g of NaOH dissolved in 90 g of water:
w/w%=1010+90×100=10%\text{w/w\%} = \frac{10}{10 + 90} \times 100 = 10\%

Volume Percentage (v/v %) and Mass by Volume (w/v %)

v/v% — used when both solute and solvent are liquids:
v/v%=volume of solutevolume of solution×100\text{v/v\%} = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100
Example: 40 mL of ethanol in 100 mL of solution → 40% v/v (the standard for alcohol labelling).

w/v% — mass of solute per 100 mL of solution. Used in medicine and pharmacy:
w/v%=mass of solute (g)volume of solution (mL)×100\text{w/v\%} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100
Example: Normal saline = 0.9 g NaCl per 100 mL = 0.9% w/v.

Mole Fraction (χ)

Mole fraction expresses concentration as a ratio of moles — temperature-independent and used extensively in Raoult's law calculations.

For a binary solution with solute (2) and solvent (1):
χ2=n2n1+n2,χ1=n1n1+n2\chi_2 = \frac{n_2}{n_1 + n_2}, \quad \chi_1 = \frac{n_1}{n_1 + n_2}

Key property: χ1+χ2=1\chi_1 + \chi_2 = 1 (mole fractions of all components always sum to 1)

Mole fraction has no units. It ranges from 0 to 1.

📖NCERT 2.1NCERT Intext

Problem

Calculate the mass percentage of benzene (CX6HX6\ce{C6H6}) and carbon tetrachloride (CClX4\ce{CCl4}) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

📖NCERT 2.2NCERT Intext

Problem

Calculate the mole fraction of benzene in a solution containing 30% by mass of benzene in carbon tetrachloride. (Molar masses: benzene = 78 g/mol, CClX4\ce{CCl4} = 154 g/mol)

JEE / NEET Exam InsightJEE / NEET
Mole fraction sum = 1 always. If χsolute=0.25\chi_{\text{solute}} = 0.25, then χsolvent=0.75\chi_{\text{solvent}} = 0.75 without calculation.
w/w% is independent of temperature (mass doesn't change with temperature). Molarity does change with temperature. This difference is a frequent exam question.
Mole fraction → Raoult's law. Always convert to mole fraction when applying Raoult's law or Henry's law — these are the only concentration units those laws use.
Quick Check

Q1.If 18 g of glucose (M = 180 g/mol) is dissolved in 90 g of water (M = 18 g/mol), what is the mole fraction of glucose?