Question

The molal elevation constant for water is 0.56 ${\text{K kg mo}}{{\text{l}}^{{\text{ - 1}}}}$. Calculate the boiling point of solution made by dissolving 6.0 g of urea ( $N{H_2}CON{H_2}$ ) in 200g of water.

Answer 1

To find the boiling point of new solution, after adding solute, we should know that there will be elevation in boiling point.
This elevation in boiling point is proportional to molality of solution. And to get the final boiling point of solution, we need to add boiling point of water with the elevation in boiling point.

Complete step by step answer:
We know one thing that, in addition to solute to pure solvent, the boiling point will be increased and freezing point will be decreased. In this case, we have been asked the new boiling point of solution.
To get boiling point of solution, we need to find first rise or elevation in boiling point, which is given as $\Delta {T_b}$
We know this elevation in boiling point is directly proportional to molality of solution.
$\Delta {T_b}\alpha {\text{ }}molality$
To remove the proportionality sign, we should bring a constant. That constant is the molal elevation constant.
$\Delta {T_b} = {K_b}m$ ---- equation 1
Molal elevation constant is given in question.
$\therefore {K_b} = 0.56Kkgmo{l^{ - 1}}$
Now, we need to find molality, which can be given by below formula:
$m = \dfrac{{{\text{moles of solute}}}}{{{\text{mass of solvent}}}}$
To find moles of solute, we need mass of solute, and molecular mass of solute. Here solute is urea.
Mass of solute = 6g
Molecular mass of urea = 60 g
${\text{moles of solute(n) = }}\dfrac{{{\text{mass of solute}}}}{{{\text{molecular mass of solute}}}}$
On substituting values,
${\text{moles of urea = }}\dfrac{6}{{60}}$
Given, mass of solvent = 200gm = 0.2kg
Now, calculate molality.
$m = \dfrac{{{\text{moles of urea}}}}{{{\text{mass of water in kg}}}}$
Substitute the values,
$m = \dfrac{{\dfrac{6}{{60}}}}{{0.2}}$
On simplification, we get molality as
$m = 0.5molek{g^{ - 1}}$
Now, to calculate elevation in boiling point, we have to substitute the molality and molal elevation constant value in equation 1, and we get
$\Delta {T_b} = 0.5 \times 0.56$
$\therefore \Delta {T_b} = 0.28K$
Boiling point of solution = boiling point of water + elevation in boiling point.
We know the boiling point of water is 100 degree Celsius.
$\therefore {\text{Boiling point of solution = 100 + 0}}{\text{.28}} = {100.28^0}C$
Hence option D is the correct one.

Note:
We should be careful in calculation of molality, where mass of solvent has to be in kg.
Take care of units, and we are also aware that the value of difference or change in temperature in the Kelvin and Celsius scale is the same.
In addition to solute to pure solvents, the properties change and these change in properties is called as colligative properties, here boiling point is changed due to addition of solute. This boiling point will increase in addition to pure solvent.