Question

A solution of an organic compound is prepared by dissolving 34.2 g in 500 g of water. Calculate the molar mass of the compound, given that ${{K}_{b}}$for water =$1.87K\,kg\,mo{{l}^{-1}}$.
Boiling point of solution = $100.104{}^\circ C$ (write up to 2 decimal places.)

Answer 1

Hint: When a non – volatile solute is added to a volatile solvent then the boiling point of solution increases. Elevation in boiling point is directly proportional to the molality, through which we can calculate molar mass. ${{K}_{b}}$is the molal elevation constant.
Formula used:
Elevation in boiling point, $\Delta {{T}_{b}}={{K}_{b}}.m$, where m is molality, and $\Delta {{T}_{b}}$is difference of initial and final boiling point.
Molality = $\dfrac{moles\,of\,solute}{Kg\,of\,solvent}$
Number of moles = $\dfrac{mass}{molar\,mass}$

Complete answer:
We have been given an amount of solute, 34.2 g which is dissolved in 500 g of water, we have to find the molar mass of this solute. Given that ${{K}_{b}}$for water =$1.87K\,kg\,mo{{l}^{-1}}$. The boiling point of solution =$100.104{}^\circ C$.
So through elevation in boiling point, $\Delta {{T}_{b}}={{K}_{b}}.m$, we will put the values and by molality formula we will find the molar mass.
Molality, m = $\dfrac{\dfrac{mass}{molar\,mass}}{Kg\,of\,solvent}$, so, m = $\dfrac{\dfrac{34.2g}{molar\,mass}}{\dfrac{500g}{100Kg}}$
Elevation in boiling point, $\Delta {{T}_{b}}$ is the difference of final boiling temperature after adding solute, and initial boiling temperature of water, as 100.104 – 100 = 0.104. Putting these values in the elevation in boiling point formula, $\Delta {{T}_{b}}={{K}_{b}}.m$ we have,
0.104 = $1.87K\,kg\,mo{{l}^{-1}}$$\times $ $\dfrac{\dfrac{34.2g}{molar\,mass}}{\dfrac{500g}{100Kg}}$
Rearranging to get molar mass, $m=\dfrac{34.2\times 2\times 1.87}{0.104}$
Molar mass, m = 1229.88 g / mole
Hence, the molar mass of the compound is 1229.88 g / mole.

Note:
The molality for any solution is always taken in kilograms, so 500 g is converted to kilograms by the factor, 1 kg = 1000 g, so 1 g = $\dfrac{1}{1000}kg$. The boiling point of water is $100{}^\circ C$, addition of a solute increases this boiling point, and hence the elevation in boiling point.