Elevation of Boiling Point Numericals

CHEColligative Properties

Adding a non-volatile solute raises the boiling point by an amount proportional to molality, allowing molar masses to be determined from boiling-point data. ISC numericals use $\Delta T_b = K_b m$ to find molar mass, molecular formula, or the elevated boiling point.

Boiling-point elevation

\Delta T_b = i K_b m = T_b(\text{solution}) - T_b(\text{solvent})

K_b

ebullioscopic constant,

m

molality,

i

van't Hoff factor

Molar mass from elevation

M_2 = \dfrac{i \, K_b \, w_2 \times 1000}{\Delta T_b \times w_1}

w_2

solute mass (g),

w_1

solvent mass (g)

Number of formula units per molecule

n = \dfrac{M_{observed}}{M_{empirical}}

e.g. sulphur as

S_8

gives

n = 8

Elevation of boiling point: $\Delta T_b = i K_b m$ , where $K_b$ is the molal (ebullioscopic) constant and $m$ is molality.
$\Delta T_b = T_b(\text{solution}) - T_b(\text{pure solvent})$ ; a degree change in $^\circ\text{C}$ equals the same change in kelvin.
Molar mass from elevation: $M = \dfrac{i K_b \times w_2 \times 1000}{\Delta T_b \times w_1}$ , with $w_2$ solute mass and $w_1$ solvent mass in grams.
Molecular formula by comparison: divide the observed molar mass by the empirical-formula mass to find the number of formula units per molecule (e.g. sulphur as $S_8$ in $CS_2$ ).
$K_b$ depends only on the solvent, not the solute; for water $K_b = 0.52\,\text{K kg mol}^{-1}$ .
For non-electrolytes $i = 1$ ; for electrolytes include $i$ so the elevation is larger than for an equimolal non-electrolyte.
A non-volatile solute lowers the solvent's vapour pressure, so the solution must be heated to a higher temperature to reach atmospheric pressure — hence the boiling point rises.
$K_b$ is the elevation produced by a $1\,\text{molal}$ solution of a non-electrolyte; it has units $\text{K kg mol}^{-1}$ and is a property of the solvent alone.
Solve numericals in order: compute $\Delta T_b$ from the two boiling points, get molality from $\Delta T_b/(iK_b)$ , then convert to moles and molar mass.
For sulphur in $CS_2$ , an observed molar mass near $256$ gives $256/32 = 8$ , confirming $S_8$ rings — a classic molecular-formula application.
Cross-check magnitude: boiling-point elevations are small (a few tenths of a degree), so an answer of several degrees for a dilute solution signals an arithmetic slip.
Trap: $m$ uses mass of solvent in kilograms — convert grams to kg, and keep solute mass in grams in the molar-mass formula.

Where the marks go

Mixing units of solvent mass — molality needs kg, but the standard molar-mass formula keeps $w_1$ in grams with the $\times 1000$ factor.
Computing $\Delta T_b$ as solvent minus solution; elevation is solution minus pure solvent (a positive quantity).
Using a solute-dependent $K_b$ ; $K_b$ belongs to the solvent only.
Dropping $i$ for an electrolyte, which underestimates the elevation and inflates the molar mass.
Forgetting to divide the observed molar mass by the atomic/empirical mass when asked for the molecular formula (e.g. number of S atoms).

How the board asks it

Numericalmolar mass from elevation, $M = \dfrac{i\,K_b\,w_2 \times 1000}{\Delta T_b \times w_1}$
$1.8\,\text{g}$ of a non-volatile, non-electrolyte solute dissolved in $90\,\text{g}$ of water raises the boiling point from $100^\circ\text{C}$ to $100.104^\circ\text{C}$ . Calculate the molar mass of the solute. (Given $K_b$ for water $= 0.52\,\text{K kg mol}^{-1}$ )
Numericalelevated boiling point from $\Delta T_b = K_b\,m$
Calculate the boiling point of a solution containing $6\,\text{g}$ of urea ( $M = 60\,\text{g mol}^{-1}$ ) dissolved in $200\,\text{g}$ of water. (Given $K_b$ for water $= 0.52\,\text{K kg mol}^{-1}$ and boiling point of pure water $= 100^\circ\text{C}$ )
Numericalevaluating $K_b$ from $\Delta T_b$ , mass and molar-mass data
When $1.2\,\text{g}$ of a non-electrolyte ( $M = 120\,\text{g mol}^{-1}$ ) is dissolved in $50\,\text{g}$ of benzene, the boiling point rises by $0.51\,\text{K}$ . Calculate the molal elevation constant $K_b$ of benzene.
Numericalvan't Hoff factor $i$ for electrolytes, $\Delta T_b = i\,K_b\,m$
A solution is prepared by dissolving $5.85\,\text{g}$ of $NaCl$ in $250\,\text{g}$ of water. Assuming complete dissociation ( $i = 2$ ), calculate the elevation in boiling point. (Given $K_b$ for water $= 0.52\,\text{K kg mol}^{-1}$ and $M_{NaCl} = 58.5\,\text{g mol}^{-1}$ )
Numericalmolecular formula by comparing observed molar mass with atomic mass
$2.56\,\text{g}$ of sulphur dissolved in $100\,\text{g}$ of $CS_2$ raises its boiling point by $0.234\,\text{K}$ . Given $K_b = 2.34\,\text{K kg mol}^{-1}$ for $CS_2$ and atomic mass of sulphur $= 32$ , find the molar mass of sulphur and hence the number of atoms in one molecule (molecular formula).

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.