Formula sheet: Solutions

Solutions12 min readLesson 1 of 8

1.1 How this sheet works The chapter is numerical, so the first page collects the relations that the rest of the pages use. Each formula is followed by its unit and the case where it applies. A symbol that appears twice carries the same meaning everywhere: n is amount in moles, m is molality, M is molarity, x is mole fraction, p is vapour pressure, K is a proportionality constant and i is the van't Hoff factor. 1.2 Concentration Mass percentage, parts per million and mole fraction are ratios and have no units. Molarity needs the volume of the solution and therefore changes when the solution is warmed. Molality uses the mass of the solvent, so it is unaffected by temperature. QuantitySymbol and unitDefined withMass percentageno unitmass of solute, mass of solutionParts per millionno unitmass of solute in 10^6 g of solutionMole fractionx, no unitmoles of one component, total molesMolarityM, mol L^-1moles of solute, volume of solutionMolalitym, mol kg^-1moles of solute, mass of solvent

Quantity	Symbol and unit	Defined with
Mass percentage	no unit	mass of solute, mass of solution
Parts per million	no unit	mass of solute in 10^6 g of solution
Mole fraction	x, no unit	moles of one component, total moles
Molarity	M, mol L^-1	moles of solute, volume of solution
Molality	m, mol kg^-1	moles of solute, mass of solvent

Mass percentage = \frac{mass of solute}{mass of solution} \times 100

Used when a solution is described by mass, as on reagent bottles

ppm = \frac{mass of solute}{mass of solution} \times 1 0^{6}

Used for very dilute solutions such as contaminants in water

x_{B} = \frac{n _{B}}{n _{A} + n _{B}}

Mole fraction of component B in a two component mixture

M = \frac{n _{solute}}{V _{solution} in litres}

Molarity, temperature dependent

m = \frac{n _{solute}}{mass of solvent in kg}

Molality, independent of temperature

1.3 Vapour pressure and gas solubility Raoult's law has two working forms. With a non-volatile solute the vapour pressure of the solution comes only from the solvent, so the relative lowering of vapour pressure equals the mole fraction of the solute. With two volatile liquids both components contribute to the total vapour pressure. Henry's law handles a gas dissolved in a liquid and uses the partial pressure of the gas above the solution.

p_{1} = x_{1} p_{1}^{0}

Raoult's law for solvent 1 over a solution that contains a non-volatile solute

\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = x_{2}

Relative lowering of vapour pressure equals the mole fraction of the non-volatile solute

p_{total} = p_{1}^{0} x_{1} + p_{2}^{0} x_{2}

Two volatile liquids that form an ideal solution

p = K_{H} x

Henry's law, with p in bar or kbar and K_H in the same pressure unit

1.4 Colligative properties Each of the four properties counts particles; what the particles are made of does not enter the equation. The equations below are written for a solute that neither dissociates nor associates. For an electrolyte or an associated solute, every one of them is multiplied by i.

Δ T_{b} = K_{b} m

Elevation of boiling point, K_b in K kg mol^-1

Δ T_{f} = K_{f} m

Depression of freezing point, K_f in K kg mol^-1

Π = C R T

Osmotic pressure, C in mol L^-1, R = 0.0821 L bar K^-1 mol^-1 if pressure is in bar

i = \frac{normal molar mass}{observed molar mass} = \frac{observed colligative property}{calculated colligative property}

van't Hoff factor

Δ T_{b} = i K_{b} m, Δ T_{f} = i K_{f} m, Π = i C R T

Variants for solutes that dissociate or associate

α = \frac{i - 1}{n - 1}

Degree of dissociation for a solute that gives n particles

α = \frac{n ( 1 - i )}{n - 1}

Degree of association for n molecules that clump together

1.5 Constants and checks R = 8.314 J K^-1 mol^-1, which is also 0.0821 L atm K^-1 mol^-1. For water, K_f = 1.86 K kg mol^-1 and K_b = 0.52 K kg mol^-1. Normal atmospheric pressure is 1.013 bar or 760 torr. Before substituting a number, convert the volume to litres for molarity, keep the solvent mass in kilograms for molality and match the pressure unit of K_H with the unit of p. A molar mass calculated from a colligative property is never the true value when the solute dissociates; the van't Hoff factor recovers it.

Exam

Examiners look for three things in a numerical answer: the formula written with symbols, the substitution with units, and the final value with the correct unit and sign. A freezing point depression is positive, while the freezing point of the solution is lower than that of the pure solvent.

Write the formula and the unit before substituting values.
Use kg of solvent in m, never kg of solution.
Multiply by i for salts, acids and bases.
Keep K_f and K_b positive; the new boiling point rises and the new freezing point falls.
Check that the final answer has the unit the question asks for.

in hinglish

Yeh Formula sheet poore chapter ka toolbox hai. Molarity moles per litre solution hoti hai aur temperature ke saath badalti hai; molality moles per kg solvent hoti hai aur temperature se free rehti hai. Henry's law mein p = K_H x, aur agar same pressure par K_H badh jaaye to solubility gir jaati hai. Raoult's law vapour pressure ko mole fraction se jodta hai; relative lowering sirf non-volatile solute ke liye x2 ke barabar hoti hai. Chaar colligative properties ke formulas yaad rakho: boiling point elevation, freezing point depression, osmotic pressure aur vapour pressure lowering. Solute dissociate ya associate kare to hamesha van't Hoff factor i lagao. Water ke liye K_f = 1.86 aur K_b = 0.52 K kg per mol, aur R = 0.0821 L atm per K mol. Numerical mein unit match karna mat bhoolo.