InterviewSolution
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InterviewSolution
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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)`. |
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Answer» 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)` |
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