Question

What is the molecular weight of sulfur if $35.5$ grams of sulfur dissolve in $100.0$ grams of $C{S_2}$ to produce a solution that has a boiling point of $49.48^\circ C$?

Answer 1

When we add a solute to a solvent the surface area of solvent gets covered by the solute particles due to which the vapour pressure decreases and hence the boiling point increases which is called boiling point elevation.
We need to look up the values of $T^\circ b$ and ${K_b}$ for . They are and ${K_b} = 2.37g/mol$
The formula for boiling point elevation is $\Delta Tb = Kbm$ where m =molality and $\Delta Tb$ = elevation in boiling point and Kb=boiling point elevation constant

Complete answer:
Given data contains,
${T_b} = 49.48^\circ C$
m of sulfur(solute) $= 35.5g$
m of $C{S_2}$ (solvent) $= 100g$
Now solving the question
$\Delta Tb{\text{ }} = Tb{\text{ }} - {\text{ }}{T^o}b = \left( {49.48{\text{ }} - {\text{ }}46.2} \right)^\circ C = {\text{ }}3.28^\circ C$
We can use this information to calculate the molality m of the solution.
$m = \dfrac{{\vartriangle {T_b}}}{{\vartriangle {K_b}}}$
We have to substitute the known values we get,
$m = \dfrac{{3.28^\circ C}}{{2.37^\circ C}}$
On simplification we get,
$m = 1.38mol/kg$
We have $100.0g$ or $0.1000kg$of $C{S_2}$. So we can calculate the moles of sulphur,
$Molality = \dfrac{{{\text{Moles of solute}}}}{{{\text{Kilograms of solvent}}}}$
Now we can substitute the known values we get,
${\text{Moles of solute = }}1.38mol/kg \times 0.1000kg$
On simplification we get,
${\text{Moles of solute}} = 0.138mol$
So $35.5g$ of sulphur $= 0.138mol$
∴ Molar mass of sulphur $= 35.5g \times 0.138mol = 35.5g/mol$
Molecular mass of sulphur is $257amu$
The molecular weight of sulfur is calculated to be $257g/mol$

Note:
We have to remember that the actual molecular mass of ${S_8}$ is $256amu$ which is very close to the value calculated for the sulfur in the question given. We have to remember that the molality is calculated based on the weight of solvent in kilograms whereas molarity is calculated based on the liters of solution. We also need to know that the value of molality does not change with change in temperature. We need to have to know that the unit of molality is mol/kg.