Question

The solution contains 7 g of solute (molar mass $ {\text{210g mo}}{{\text{l}}^{{\text{ - 1}}}} $ ) in 350g of acetone raised the boiling point of acetone from $ {\text{5}}{{\text{6}}^{\text{0}}}{\text{C to 56}}{\text{.}}{{\text{3}}^{\text{0}}}{\text{C}} $ . The value of ebullioscopic constant of acetone in $ {\text{K}}{\text{.Kgmo}}{{\text{l}}^{{\text{ - 1}}}} $ is:
(A) $ 2.66 $
(B) $ 3.15 $
(C) $ 4.12 $
(D) $ 2.86 $

Answer 1

Hint: We need to find out the molality. Then further ebullioscopic constants can be found out using change in temperature. We shall calculate the molality from the formula given and then, substitute the value in the elevation in boiling point formula and calculate the value of ebullioscopic constant.

Formula used: $ {\text{Molality(m) = }}\dfrac{{{\text{No of moles of solute}}}}{{{\text{Kg of solvent}}}} $
 $ \Delta {T_b} = {K_b}.m $
Here, we have to find $ {K_b} $ .

Complete step by step solution:
We already know the boiling point of acetone.
 $ {T_b} = {56.3^0}C $ = $ \;\left( {273 + 56.3} \right){\text{ }}K{\text{ }} = 329.3K $
 $ {T_b}^0 = {56^0}C $ = $ \left( {273 + 56} \right){\text{ }}K{\text{ }} = 329K $
Mass of solute = 7g
Mass of solution = $ \left( {350 + 7} \right){\text{ g }} = {\text{ }}357{\text{g}} $
 $ {\text{Molality(m) = }}\dfrac{{{\text{No of moles of solute}}}}{{{\text{Kg of solvent}}}} $
 $ {\text{m = }}\dfrac{{\left( {\dfrac{{{\text{mass of solute}}}}{{{\text{molar mass of solute}}}}} \right)}}{{{\text{Kg of solvent}}}} $
On substituting the values,
 $ m = \dfrac{{\left( {\dfrac{7}{{210}}} \right)}}{{\left( {\dfrac{{350}}{{1000}}} \right)}} = \dfrac{{10}}{{105}} $
As we know that, depression in boiling point,
 $ \Delta {T_b} = {K_b}.m $
where,
 $ {T_b} - {T_b}^0 = 329.3 - 329 = 0.3K $
 $ {K_b} = $ ebullioscopic constant of acetone
 $ \therefore 0.3 = {K_b} \times \dfrac{{105}}{{10}} $
 $ {K_b} = 3.15{\text{KKgmo}}{{\text{l}}^{{\text{ - 1}}}} $
​So, we need to see from the above options, and select the correct option.
Thus, the correct answer is option B.

Note:
We should keep in mind that Ebullioscopic constant is the constant that expresses the amount by which the boiling point of a solvent is raised by a non-dissociating solute. In thermodynamics, the ebullioscopic constant relates molality b to boiling point elevation. Through the procedure called ebullioscopy, a known constant can be used to calculate an unknown molar mass. This property of elevation of boiling point is a colligative property. Colligative properties are important properties of solutions as they describe how the properties of the solvent will change as solute (or solutes) is (are) added. It means that the property, in this case change in temperature, depends on the number of particles dissolved into the solvent and not the nature of those particles. Cryoscopic and ebullioscopic constants are generally tabulated using molality as the unit of solute concentration rather than mole fraction. In this form, the equation for calculating the magnitude of the freezing point decrease or the boiling point increase.