Question

: Elevation of boiling point of 1 molar aqueous glucose solution (density of the solvent $1.2g/ml$) is
A.${K_b}$
B.$1.2{K_b}$
C.$1.30{K_b}$
D.$0.833{K_b}$

Answer 1

Colligative properties: It is defined as the ratio of number of solute particles to the number of solvent particles. It only depends on the number of ions and not on the nature of the solute.
Elevation in boiling point: It is defined as an increase in boiling point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.

Complete step by step solution:
There are four colligative properties: Depression in freezing point, elevation in boiling point, osmotic pressure and lowering in vapour pressure.
Depression in freezing point: It is defined as the decrease in freezing point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.
Elevation in boiling point: It is defined as an increase in boiling point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.
Osmotic pressure: It is defined as the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.
Relative lowering in vapour pressure: When non- volatile solutes are added to a solution then there is a decrease in its vapour pressure, which is known as relative lowering in vapour pressure.
Here we are given with the density i.e. $1.2g/ml = 1.2kg/L$ and also we are given that the concentration of the solution is $1molar$.
Hence the mass of the solution will be density multiplied with its volume here the volume is $1L$ and density is $1.2Kg/L$. So the value of the mass of the solution will be $1.2Kg$. Now here glucose acts as a solute to the solution and the molar mass of glucose is $180gm$. So the mass of the solvent will be$1200 - 180 = 1020gm = 1.02Kg$.
Now molality is the ratio of moles of solute to the mass of solvent in Kg.
Molality $= \dfrac{1}{{1.02}}$ because here the number of moles of solute is $1$ and mass of solvent is $1.02$. So molality is equal to $0.98$.
Also the elevation in boiling point is the product of molality with molal elevation constant.
So elevation in boiling point is $0.98{K_b}$ which is approximately equal to $1{K_b}$.

Hence, option A is correct.

Note:
Van’t Hoff factor: It is defined as the ratio of actual number of particles in the solution after dissociation or association to the number of particles for which no ionization takes place. It is represented by $i$.