Question

The latent heat of vaporisation of water is $9700cal/mole$ and if the boiling point is $100$, the ebullioscopic constant of water is :
A. $0.516$
B. $1.026$
C. $10.26$
D. $1.832$

Answer 1

Hint: The ebullioscopic constant of water relies on the molar mass of the solute in the solution. This constant relates with the molality of a solution to the increase in the boiling point which is directly proportional to the molality of the solution.

Formula Used:
We can find the ebullioscopic constant of water by using the formula :
${{K}_{b}}=\dfrac{R{{T}^{2}}_{b}M}{\Delta H\times 1000}$
Where R is the universal gas constant,
${{T}_{b}}$is the boiling point of the solvent, M is the molecular weight of the solvent,
$\Delta {{H}_{vap}}$is the enthalpy of vapourization.

Complete Step by Step Solution:
In this question, we have to find the ebullioscopic constant of water.
Here ${{T}_{b}}$ = $100+273 = 373K$
And $\Delta {{H}_{vap}}$ = $\dfrac{9700}{18}cal/gm$
And we know the formula of ebullioscopic constant of water is
${{K}_{b}} =\dfrac{R{{T}^{2}}_{b}M}{\Delta H\times 1000}$
Now we put the values in the above equation, we get
${{K}_{b}}=\dfrac{2\times {{(373)}^{2}}\times 18}{9700\times 1000}$
Solving the above equation, we get
${{K}_{b}}=0.516$
Hence the ebullioscopic constant of water is 0.516
Thus, Option (A) is correct.

Note: The properties that are independent of the nature of solute present in the solution, but not only from the number of solute particles but also from the moles of solute particles are known as colligative properties. There are four colligative properties which are : the lowering of vapour pressure, the depression of freezing point, the elevation in the boiling point, and the osmotic pressure.