Choose the denominator first: solution volume, solvent mass, or total moles.
JEE Main · JEE Advanced · NEET UG43 min study guide9 indexed sections
BY THE END, YOU CAN
Distinguish true solutions from suspensions and colloids, and classify solution phases.
Convert among molarity, molality, mole fraction and mass-based concentration.
Use density without confusing solution mass with solvent mass.
Apply dilution and mixing balances only when their conservation assumptions hold.
Interpret saturation, temperature dependence and electrolyte analytical concentration.
BEFORE YOU START
Mole concept and molar mass.
Mass, volume and density units.
Balanced reaction equations and basic algebra.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A solution is a homogeneous phase, not necessarily a liquid
A solution has components dispersed uniformly on the molecular or ionic scale within one phase. Water is a common solvent, but it is not part of the definition. Air can be treated as a gaseous solution; ethanol and water can form a liquid solution; a single-phase copper–nickel alloy is a solid solution. A material’s everyday appearance is not enough to prove molecular homogeneity.
The solvent generally sets the phase of the solution and is often the more abundant component. In gas mixtures or nearly equal liquid mixtures, the naming choice may be conventional. Solute and solvent are roles in a mixture, not permanent properties of a substance: water can be the solvent in dilute aqueous ethanol or the minority component in ethanol-rich material.
Examples organized by final phase are more useful than memorizing a nine-cell table without conditions. In gaseous solutions, components are gases after mixing; humid air contains molecular water vapor, whereas fog contains liquid droplets and is not a single gas solution. Liquid solutions can contain dissolved gases, liquids or solids. A solid solution can contain substitutional atoms or small interstitial atoms, provided the relevant material is one homogeneous solid phase.
Not every alloy is a single solid solution, and not every apparently uniform dispersion is a true solution. A suspension contains larger dispersed particles that may settle. A colloid has dispersed structures larger than ordinary solute molecules and can scatter light appreciably. Milk, smoke and fog therefore require a dispersed-phase description rather than the simple molecular-solution model.
Two liquids can be miscible over the composition range or separate into two phases. A mixture with two visibly separated liquid layers is not one solution with an average concentration, although each layer may itself be a solution. The amount of each phase and its composition must then be specified separately. All concentration formulas in this lesson refer to a clearly identified homogeneous sample or phase.
ORIGINAL WORKED EXAMPLE
Classify the phase before calculating
Classify humid air without droplets, fog, and a homogeneous single-phase copper–nickel alloy. Which can be modeled as true solutions?
Humid air contains gas molecules in one gaseous phase and can be treated as a gas solution.
Fog contains liquid droplets dispersed in gas and is a colloidal aerosol, not a single gas solution.
The stated single-phase alloy has atoms mixed in one solid phase and is a solid solution.
The explicit single-phase condition matters: an arbitrary alloy can contain multiple phases.
Answer Humid air and the stated alloy are solutions; fog is a dispersed liquid-in-gas colloid.
02 / UNDERSTAND & APPLY
2. Concentrated and saturated answer different questions
Concentration states how much of a specified component is present relative to a chosen amount of mixture or solvent. Saturation asks whether the solution is at equilibrium with an additional phase of that solute under the stated conditions. A sparingly soluble salt may form a saturated solution whose concentration is very low, whereas a highly soluble substance can have a concentrated but unsaturated solution.
An unsaturated solution can dissolve more solute under the stated conditions. A saturated solution is at its solubility limit in equilibrium with the relevant solute phase. A supersaturated solution contains more dissolved solute than the equilibrium value and is metastable; suitable nucleation can trigger crystallization. These descriptions require temperature, pressure where relevant, solvent and chemical conditions.
The undissolved solid in a bottle of saturated solution is not included in the mass or volume of a clear aliquot withdrawn from the liquid. Including it in a concentration denominator or numerator changes the sample being described. Likewise, a mass balance for the entire bottle and a concentration calculation for its liquid phase are different tasks.
Dissolution does not always raise temperature, and greater temperature does not always increase solubility. Mixing is governed by competing energetic and entropy contributions. For the present calculations, use the specified solubility data rather than a universal temperature slogan. If precipitation, reaction, evaporation or phase separation occurs, the dissolved amount must be found before calculating concentration.
Concentration is also species-specific. A solution prepared from a given amount of salt has a formal or analytical formula-unit concentration. After dissociation, individual ion concentrations may be different multiples of it; after hydrolysis or complexation, their free concentrations may differ further. State whether a number refers to added compound, total analytical element, or free chemical species.
03 / UNDERSTAND & APPLY
3. Molarity counts moles per final solution volume
Molarity C, often written M in classroom notation, is n_solute/V_solution with volume in liters. The denominator is the volume of the finished solution, not the volume of solvent used before dissolving. Dissolving a solid in 250 mL of water does not establish that the final solution volume is exactly 250 mL.
To prepare a specified volume from a solid in the ideal quantitative calculation, find n=CV and then mass=n times molar mass. A practical volumetric preparation dissolves the material in less than the final solvent amount and then brings the equilibrated solution to the final calibration mark. Purity, hydration state and quantitative transfer matter when converting a weighed material into analytical moles.
An aliquot of a homogeneous solution has the same molarity as the bulk sample, but fewer total moles. A 10 mL portion is not more concentrated than a 100 mL portion from the same well-mixed solution. Amount and concentration are linked by n=CV and must not be treated as interchangeable quantities.
Molarity can change with temperature because liquid volume changes even when solute moles remain constant. Statements comparing molarities should therefore identify the temperature or assume it is fixed. In ordinary dilute classroom problems the effect may be neglected, but that approximation does not make molarity fundamentally temperature-independent.
C=n_solute/V_solution
C in mol L⁻¹ when n is in mol and final solution volume is in L.
ORIGINAL WORKED EXAMPLE
Prepare a target analytical concentration
How much anhydrous Na₂SO₄, molar mass 142.0 g mol⁻¹, is needed for 250.0 mL of 0.1200 M solution, assuming pure material and complete transfer?
Convert the final volume to 0.2500 L.
Required formula-unit amount is 0.1200×0.2500=0.03000 mol.
Mass is 0.03000×142.0=4.260 g.
The final solution is made up to 250.0 mL; this is not an instruction to add the solid to exactly 250.0 mL of water.
Answer 4.260 g of anhydrous Na₂SO₄.
04 / UNDERSTAND & APPLY
4. Molality and mole fraction use different conserved denominators
Molality b, frequently denoted m, is moles of solute per kilogram of solvent. The solvent mass excludes the solute. If 12 g of solute is dissolved in 88 g of water, the denominator is 0.088 kg, not 0.100 kg. Molality is useful for colligative-property formulas because the solvent-based composition remains unchanged by simple thermal expansion.
Mole fraction x_i is n_i divided by the sum of the moles of all components in the specified mixture. Each x_i is dimensionless and the fractions sum to one. A mass percentage cannot be substituted for a mole fraction unless a justified conversion has been made; unequal molar masses make the corresponding fractions different.
For a binary nonelectrolyte solution described in terms of neutral components, choose 1 kg of solvent to connect the definitions. Then n_solute=b and n_solvent=1000/M_solvent when molar mass is in g mol⁻¹. Therefore x_solute=b/[b+1000/M_solvent]. This relation does not need the solution density.
Mass-based and mole-based measures remain constant under a temperature change only if composition is unchanged. Evaporation, reaction or precipitation can change them. For electrolytes, distinguish component mole fractions based on formula amounts from microscopic species fractions after dissociation. A problem about particle effects may require a speciation or van’t Hoff model beyond a simple formula-unit fraction.
b=n_solute/(mass_solvent in kg); x_i=n_i/Σn_j
Molality uses solvent mass; mole fraction uses total moles of the specified components.
ORIGINAL WORKED EXAMPLE
Two concentrations from one weighed mixture
A nonelectrolyte of molar mass 60.0 g mol⁻¹ has mass 12.0 g and is mixed with 180.0 g water, taken as 18.0 g mol⁻¹. Find molality and solute mole fraction.
Solute amount is 12.0/60.0=0.200 mol.
Solvent mass is 0.1800 kg, so b=0.200/0.1800=1.111 mol kg⁻¹.
Water amount is 180.0/18.0=10.00 mol.
x_solute=0.200/(10.00+0.200)=0.01961.
Answer b≈1.11 mol kg⁻¹; x_solute≈0.0196.
05 / UNDERSTAND & APPLY
5. Percent, ppm and ppb need an explicit basis
Mass percent is 100 times solute mass divided by total solution mass. A 12% by-mass solution contains 12 g solute in 100 g solution, leaving 88 g solvent in a binary mixture. It does not mean 12 g solute plus 100 g solvent. Choosing a 100 g solution basis often makes percentage conversions transparent.
Volume percent is commonly stated as a component volume per final solution volume, with the preparation convention and temperature specified. Volume contraction on mixing means that component volumes need not add to the final volume. Mass/volume percent, another distinct convention, usually means grams of solute per 100 mL solution. The bare symbol % is incomplete when the basis is not clear.
For mass fractions, ppm means a mass ratio multiplied by one million and ppb by one billion. Thus 1 ppm by mass is 1 mg per kg of solution. It equals approximately 1 mg per liter only when the solution density is approximately 1 kg L⁻¹. Concentrated, organic or dense solutions need the actual density conversion.
Gas composition can instead be expressed in mole-fraction or volume-fraction ppm. Under an ideal-gas model these two fractions coincide at common conditions, but neither is automatically a mass fraction. Carry the basis with the unit whenever there is ambiguity. “Five ppm” without context can conceal a significant conversion error.
mass %=100 m_solute/m_solution; ppm_mass=10⁶ m_solute/m_solution
Use the same mass unit in numerator and denominator.
ORIGINAL WORKED EXAMPLE
Trace mass concentration with non-unit density
A solution contains 8.00 ppm solute by mass and has density 1.20 kg L⁻¹. Find its solute concentration in mg L⁻¹.
8.00 ppm by mass means 8.00 mg per kg solution.
One liter of this solution has mass 1.20 kg.
The solute amount per liter is 8.00×1.20=9.60 mg.
Replacing ppm directly by mg L⁻¹ would incorrectly assume unit density.
Answer 9.60 mg L⁻¹.
06 / UNDERSTAND & APPLY
6. Density connects volume-based and mass-based measures
For a binary solution with molarity C, density ρ in g mL⁻¹ and solute molar mass M_s in g mol⁻¹, take exactly 1 L of solution as a basis. Its mass is 1000ρ g. It contains C moles of solute with mass CM_s g, leaving 1000ρ−CM_s grams of solvent. Dividing C by that solvent mass in kilograms gives the molality.
The resulting relation is b=1000C/(1000ρ−CM_s). Its denominator must be positive. This positivity check can reveal an impossible density–concentration combination or inconsistent units. The reverse relation is C=1000ρb/(1000+bM_s), which follows by choosing 1 kg solvent and converting total solution mass to volume.
If mass fraction w is known, a 1 L basis gives solute mass 1000ρw g and therefore C=1000ρw/M_s. With a numerical mass percentage p, w=p/100, so C=10ρp/M_s. These formulas assume the stated density belongs to that solution at the relevant temperature, not to the pure solvent.
Do not memorize a conversion formula without its unit convention. Substituting density in kg m⁻³ into a formula expecting g mL⁻¹ introduces a factor of one thousand. Rebuilding the calculation from a convenient physical basis is usually safer than trying to recall every rearranged expression.
b=1000C/(1000ρ−CM_s)
C in mol L⁻¹, ρ in g mL⁻¹, M_s in g mol⁻¹; binary solution.
C=1000ρw/M_s
w is mass fraction, not the numerical percentage.
ORIGINAL WORKED EXAMPLE
Convert a labeled mass percentage
A binary aqueous solution is 20.0% by mass in a solute of molar mass 40.0 g mol⁻¹ and has density 1.15 g mL⁻¹. Find molarity and molality.
Choose 100.0 g solution: solute mass 20.0 g and water mass 80.0 g.
Solute amount is 20.0/40.0=0.500 mol.
Solution volume is 100.0/1.15=86.96 mL=0.08696 L.
C=0.500/0.08696=5.75 M; b=0.500/0.0800=6.25 mol kg⁻¹.
Answer C=5.75 M and b=6.25 mol kg⁻¹.
07 / UNDERSTAND & APPLY
7. Dilution conserves solute amount, not concentration
When solvent is added without losing, reacting or precipitating solute, its analytical amount stays constant. Therefore C_initial V_initial=C_final V_final. The final volume is the total solution volume after dilution. It is not merely the volume of solvent added. Both volume units may cancel if they are identical, but the concentration units must also be consistent.
For mixing two nonreacting solutions containing the same analytical solute, add their amounts: n_total=C₁V₁+C₂V₂. Divide by the actual final mixture volume. Only when volume additivity is stated or justified may that denominator be replaced by V₁+V₂. Masses are additive in a closed mixing operation; liquid volumes need not be.
If the solutions react, carry out stoichiometry before calculating remaining species concentrations. The sum of initial acid moles and base moles is not the final amount of either species. Similarly, a precipitate removes dissolved material while the total element mass is conserved. An analytical conservation equation and a dissolved-species concentration are not always the same quantity.
Serial dilution multiplies dilution factors. Transferring one tenth of a well-mixed solution and diluting that aliquot to ten times its volume lowers concentration by a factor of ten; repeating the operation lowers the original concentration by a factor of one hundred. The amount remaining in the original flask is irrelevant to the concentration of the selected aliquot.
ORIGINAL WORKED EXAMPLE
Dilute a stock solution
What stock volume of 1.50 M solution is needed to make 300.0 mL of 0.200 M solution, assuming no reaction or loss?
Required amount is 0.200×0.3000=0.0600 mol.
Stock volume is 0.0600/1.50=0.0400 L.
Transfer 40.0 mL stock and dilute to a final volume of 300.0 mL.
Do not interpret the result as adding exactly 300.0 mL solvent to the stock.
Answer 40.0 mL stock, diluted to 300.0 mL total.
ORIGINAL WORKED EXAMPLE
Mix analytical amounts
Mix 100.0 mL of 0.300 M and 150.0 mL of 0.100 M solutions of the same nonreacting solute. Assume additive volumes. Find the final concentration.
First portion contains 0.1000×0.300=0.0300 mol.
Second portion contains 0.1500×0.100=0.0150 mol.
Total amount is 0.0450 mol and volume is 0.2500 L.
C_final=0.0450/0.2500=0.180 M; it lies between the two initial concentrations.
Answer 0.180 M.
08 / UNDERSTAND & APPLY
8. Name the chemical quantity before choosing a formula
If a 0.100 M CaCl₂ solution is modeled as completely dissociated with negligible complexation, its analytical calcium concentration is 0.100 M and chloride concentration is 0.200 M. The total ideal solute-particle concentration is 0.300 M. These are three different quantities derived from the same prepared solution.
For weak electrolytes, hydrolyzing salts and metal complexes, free species concentrations require equilibrium calculations. A weak acid prepared at 0.100 M does not generally have [H⁺]=0.100 M. Conversely, the total analytical acid concentration can remain 0.100 M even while the distribution between HA and A⁻ changes.
Normality is an equivalent concentration and depends on the specified reaction. A mole of a polyprotic acid can supply different numbers of equivalents depending on which neutralization endpoint is intended. Redox equivalent factors likewise depend on the actual oxidation-state change. Molarity is therefore a less ambiguous starting measure when the reaction is not yet specified.
A robust workflow is to identify the phase and named species, choose a convenient mass or volume basis, convert to moles, calculate the correct denominator, and check units and limiting behavior. A fraction outside zero to one, negative solvent mass, or a nonreacting mixture concentration outside the initial range under an additive-volume assumption indicates a premise or arithmetic error worth investigating.
Using solvent volume for molarity or solution mass for molality.
Treating every percent as mass percent or every ppm as mg L⁻¹.
Assuming volumes add without an explicit model.
Applying dilution conservation across a chemical reaction without a species balance.
Calling a saturated solution necessarily concentrated.
Equating analytical salt concentration with every ion concentration.
09 / UNDERSTAND & APPLY
9. Solve with the denominator visible
For each exercise, write what the numerator counts and what the denominator measures. A correct numerical answer with the wrong physical denominator will not transfer reliably to a different problem.
ORIGINAL PRACTICE · 1
A sample contains 0.0600 mol solute in 400.0 mL final solution. Find molarity.
Reveal answer & reasoning
0.150 M.
Convert 400.0 mL to 0.4000 L, then C=0.0600/0.4000=0.150 mol L⁻¹.
ORIGINAL PRACTICE · 2
Dissolve 0.250 mol solute in 500.0 g solvent. Find molality.
Reveal answer & reasoning
0.500 mol kg⁻¹.
Solvent mass is 0.5000 kg, so b=0.250/0.5000. No solution volume is needed.
ORIGINAL PRACTICE · 3
A binary mixture contains 2.00 mol A and 3.00 mol B. Find both mole fractions.
Reveal answer & reasoning
x_A=0.400 and x_B=0.600.
Total amount is 5.00 mol. Divide each component amount by 5.00; the two fractions sum to one.
ORIGINAL PRACTICE · 4
A solution has 15.0 g solute and 85.0 g solvent. Find mass percent.
Reveal answer & reasoning
15.0% by mass.
Total solution mass is 100.0 g, and 100×15.0/100.0=15.0. The denominator is not 85.0 g.
ORIGINAL PRACTICE · 5
A 2.00 ppm by-mass solution has density 0.800 kg L⁻¹. Find mg L⁻¹.
Reveal answer & reasoning
1.60 mg L⁻¹.
2.00 mg per kg multiplied by 0.800 kg per liter gives 1.60 mg per liter.
ORIGINAL PRACTICE · 6
A 25.0 mL aliquot of 0.400 M stock is diluted to 200.0 mL. Find final molarity.
Reveal answer & reasoning
0.0500 M.
Conserved amount gives C_f=0.400×25.0/200.0. Both volumes use the same unit, so their ratio is dimensionless.
ORIGINAL PRACTICE · 7
A 0.0500 M AlCl₃ solution is completely dissociated with no hydrolysis or complexation in the stated model. Find chloride concentration.
Reveal answer & reasoning
0.150 M.
Each formula unit yields three chloride ions, so [Cl⁻]=3×0.0500. The ideal-model qualification avoids confusing analytical and free-species behavior.
ORIGINAL PRACTICE · 8
A sparingly soluble compound has a very dilute solution in equilibrium with excess solid. Is the solution unsaturated because it is dilute?
Reveal answer & reasoning
No. It is saturated under those equilibrium conditions.
Dilute describes concentration; saturated describes equilibrium with the solute phase. Low solubility permits both descriptions to apply simultaneously.
KEEP THE BIG IDEAS
Quick revision
Define the phase, solute and denominator before converting.
Molarity uses final solution volume; molality uses solvent mass.
Mole fractions sum to one for the specified set of components.
Density conversions require solution density at the relevant conditions.
Dilution and mixing equations follow material conservation, with reaction and phase changes treated separately.
Saturation and concentration are independent descriptions.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.