when `1.80 g` of a nonvolatile solute is dissolved in `90 g` of benzene, the boiling point is raised to `354.11K`. If the boiling point of benzene is `353.23K` and `K_(b)` for benzene is `2.53 KKg mol^(-1)`, calculate the molecular mass of the solute.
Strategy: From the boiling point of the solution, calculate the boiling point elevation, `DeltaT_(b)`, then solve the equation `DeltaT_(b)=K_(b)m` for the molality `m`. Molality equals moles of solute divided by kilograms of solvent (benzene). By substituting values for molality and kilograms `C_(6)H_(6)`, we can solve for moles of solute. The molar mass of solute equals mass of solute `(1.80 g)` divided by moles of solute. The molecular mass (in amu) has the same numerical value as molar mass in `g mol^(-1)`.

Explanation

Step 1. Calculate the molality
The boiling point elevation is
`DeltaT_(b)=(T_(b)-T_(b)^(0))=(354.11- 353.23)K`
=0.88 K
So the molality of the solution is
`m=(DeltaT_(b))/K_(b)=(0.88 K)/(2.53 K Kg mol^(-1))
=0.348 mol Kg^(-1)`
Step 2. Calculate the moles of solute
The molality of an aqueous solution is
`Molaity=("moles of solute")/("Kg benzene")`
Hence, for given solution
`0.348 mol Kg^(-1)=("moles of solute")/(90xx10^(-3)Kg)`
Rearranging this equation, we get
Moles of solute `=(0.348 mol kg^(-1))(90xx10^(-3) Kg)`
`=3.132xx10^(-2) mol`
Step 3. Calculate the molar mass of solute
Molar mass=`(mass)_(solute)/("moles"_(solute))`
=`(1.80 g)/(3.132xx10^(-2) mol)`
=`57.5 g mol^(-1)`
Thus, the molecular mass is 57.5 amu.
Alternatively, we can use Equation (2.65) to get the molar mass
`M_(2)(g mol^(-1))=(K_(b)(K Kg mol^(-1))(w_(2) g)(1000 g Kg^(-1)))/(DeltaT_(b)(K)(w_(1)g))`
`M_(2)=((2.53 K Kg mol^(-1))(1.8 g)(1000 g Kg^(-1)))/((0.88 K)(90 g))`
=`57.5 g mol^(-1)`