At `TK`, the vapor pressure of pure benzene is `750 mmHg` and that of a solution obtained by dissolving a nonvolatile solute is `700 mmHg`. Estimate the molarity of the solution.
Strategy: Using `2nd` Raoult's law, find the mole fraction of ninvolatile solute. Through the knowledge of mole fraction, we can find the moles of solute of solvent. Finally we can convert the moles of solvent into grams of solvent to calculate the molality.

Explanation

Step 1. Calculate the mole fraction of solute according to equation `(2.55)`, we have
`(DeltaP_(1))/P_(1)^(0)=chi_(2)`
`((750-700) mmHg)/(750 mmHg)=chi_(2)`
or `chi_(2)=1/15`
Step 2. Calculate moles of solute and solvent
By definition
`chi_(2)=n_(solute)/n_("Total")=1/15`
Since solution is a homogeneous mixture of solute and solvent, we can consider any size of solution to calculate its molality. Thus, if we consider 15 moles of solution, then
`n_("solute")=1`
`n_("solvent")=15-1=14`
step 3. Calculate mass of solvent
Converting moles into grams gives the mass of solvent, benzene (molecular mass, `78 u)`
`n_("solv")=(mass_("solv"))/(molar mass_("solv"))`
or `mass_("solv")=(n_("solv"))(molar mass_("solv"))`
`=(14 mol)(78 g mol^(-1))`
`=1008 g`
Step 4. Calculate molality
By definition
Molality (m)`=n_(solute)/g_("solv")xx(1000 g)/(kg)`
`=(1 mol)/(1008 g)xx(1000 g)/(kg)`
`=0.9 mol kg^(-1)`
Alternatively, we can proceed as follows:
By definition, molality (m) is the number of moles of solute per kg of solvent. Thus, if we consider just 1 kg of benzene (solvent), then number of moles of solute `(n_(2))` will be equal to the molality of solution.
`m=n_(solute)/(kg_("solv"))`
For `1 kg` of solvent, we have
`m=n_(solute)`
By definition
`chi_(2)=n_(2)/(n_(1)+n_(2))`
where `n_(1)` and `n_(2)` are the numbers of moles of solvent and solute, respectively. Sonce
`n_(2)=m`
and the number of moles of benzene `(n_(1))` in 1 kg of benzene is
`n_(1)=("mass(benzene)")/("molar mass (banzene)")`
`=(1000 g)/(78 g mol^(-1))`
`=12.82 mol` benzene
Substituting these result into the definition of mole fraction, we have
`1/15=m/(12.82+m)`
`12.82+m=15 m`
`12.82=14 m`
or `m=12.82/14`
`=0.9 mol kg^(-1)`