RLVP and Molar Mass Determination

CHEColligative Properties

Relative lowering of vapour pressure (RLVP) equals the mole fraction of the solute, giving a direct route to a solute's molar mass from vapour-pressure data. ISC numericals compute the solute mole fraction first, then its molar mass.

Relative lowering of vapour pressure

\dfrac{p^\circ - p_s}{p^\circ} = x_{solute} = \dfrac{n_2}{n_1 + n_2}

subscript 2 = solute, 1 = solvent;

p^\circ

pure-solvent vapour pressure

Dilute-solution approximation

\dfrac{p^\circ - p_s}{p^\circ} \approx \dfrac{n_2}{n_1} = \dfrac{w_2 / M_2}{w_1 / M_1}

valid when

n_2 \ll n_1

Molar mass from RLVP

M_2 = \dfrac{w_2 \, M_1 \, p^\circ}{w_1 \,(p^\circ - p_s)}

w_2

solute mass,

w_1

solvent mass,

M_1

solvent molar mass

RLVP: $\dfrac{p^\circ - p_s}{p^\circ} = x_{solute} = \dfrac{n_2}{n_1 + n_2}$ , where subscript 2 is solute and 1 is solvent.
For dilute solutions $n_2 \ll n_1$ , so the relation simplifies to $\dfrac{p^\circ - p_s}{p^\circ} \approx \dfrac{n_2}{n_1} = \dfrac{w_2 / M_2}{w_1 / M_1}$ .
Molar mass from RLVP: $M_2 = \dfrac{w_2 \, M_1 \, p^\circ}{w_1 \,(p^\circ - p_s)}$ using the dilute approximation.
RLVP is a colligative property — it depends on the number of solute particles, so include $i$ for electrolytes.
Steps: compute $\dfrac{p^\circ - p_s}{p^\circ}$ to get $x_{solute}$ , find moles of solvent, then solve for moles (hence molar mass) of solute.
RLVP is dimensionless because it is a ratio of pressures, so the answer is independent of the pressure unit (mmHg, atm or Pa) as long as $p^\circ$ and $p_s$ share it.
The exact form $\dfrac{p^\circ - p_s}{p^\circ} = \dfrac{n_2}{n_1 + n_2}$ should be used when the solution is concentrated; the $n_2/n_1$ shortcut is only the dilute limit.
For an electrolyte the number of particles becomes $i\,n_2$ , so RLVP $= \dfrac{i\,n_2}{n_1 + i\,n_2}$ and the apparent molar mass comes out abnormally low.
RLVP was historically the basis of the Ostwald-Walker dynamic method for measuring vapour-pressure lowering and molar mass.
Cross-link: $p^\circ - p_s = p^\circ x_{solute}$ , so vapour-pressure lowering itself is proportional to the solute mole fraction (a consequence of Raoult's law).
Trap: $p^\circ$ is the vapour pressure of the pure solvent, and the lowering must be expressed relative to $p^\circ$ , not to $p_s$ .

Where the marks go

Dividing the lowering by $p_s$ instead of $p^\circ$ — RLVP is always relative to the pure-solvent vapour pressure.
Using the dilute approximation $n_2/n_1$ when the solution is concentrated; use the exact $n_2/(n_1+n_2)$ form instead.
Omitting the van't Hoff factor for electrolytes, which makes the computed molar mass abnormally low.
Using moles of solution rather than moles of solvent ( $n_1$ ) in the denominator of the mole-fraction expression.
Carrying inconsistent pressure units for $p^\circ$ and $p_s$ — the ratio only cancels when both use the same unit.

How the board asks it

Numericalmolar mass from rlvp
The vapour pressure of pure benzene at a certain temperature is $640\,\text{mmHg}$ . A non-volatile, non-electrolyte solid weighing $2.175\,\text{g}$ is added to $39.0\,\text{g}$ of benzene ( $M_1 = 78\,\text{g mol}^{-1}$ ); the vapour pressure of the solution falls to $600\,\text{mmHg}$ . Calculate the molar mass of the solid.
Numericalrelative lowering of vapour pressure
$18\,\text{g}$ of glucose ( $M = 180\,\text{g mol}^{-1}$ ) is dissolved in $178.2\,\text{g}$ of water at $100^\circ\text{C}$ . If the vapour pressure of pure water at this temperature is $760\,\text{mmHg}$ , calculate the relative lowering of vapour pressure and the vapour pressure of the solution.
Derive / proverlvp equals solute mole fraction
Starting from Raoult's law, derive the relation $M_2 = \dfrac{w_2\,M_1\,p^\circ}{w_1\,(p^\circ - p_s)}$ for the molar mass of a non-volatile solute, clearly stating the dilute-solution approximation $n_2 \ll n_1$ used.
Define / stateraoult's law statement
State Raoult's law for a solution containing a non-volatile solute and show that $\dfrac{p^\circ - p_s}{p^\circ} = x_2$ , the mole fraction of the solute.
Numericali n_2 for electrolytes
The vapour pressure of water is $17.5\,\text{mmHg}$ at $20^\circ\text{C}$ . Calculate the vapour pressure of a solution of $5.85\,\text{g}$ of $NaCl$ in $100\,\text{g}$ of water, assuming $NaCl$ is completely dissociated ( $i = 2$ ).

Practise this topic

PreviousIdeal vs Non-Ideal Solutions and Azeotropes NextElevation of Boiling Point Numericals

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.