Question

If for a sucrose solution elevation in boiling point is \[{\rm{0}}{\rm{.1^\circ C}}\] then what will be the boiling point of NaCl solution for the same molal concentration
A. \[{\rm{0}}{\rm{.1^\circ C}}\]
B. \[{\rm{0}}{\rm{.2^\circ C}}\]
C. \[{\rm{0}}{\rm{.08^\circ C}}\]
D. \[{\rm{0}}{\rm{.01^\circ C}}\]

Answer 1

Hint: Elevation in boiling point is a process by which a solution maintains a higher boiling point than a pure solvent. This occurs when a non-volatile solute is put into a pure solvent.

Formula Used:
\[{\rm{i = }}\frac{{{\rm{Total number of moles of particles after association/dissociation}}}}{{{\rm{Number of moles of particles before association/dissociation}}}}\] where I= van't Hoff factor

Complete Step by Step Solution:
In the given question, the elevation in boiling point for sucrose solution is \[{\rm{0}}{\rm{.1^\circ C}}\].
We have to find out the elevation in boiling point for NaCl for the same molal concentration.
We know that ionic compounds undergo dissociation in water.
They dissociate into positive ions called cations and negative ions called anions.
If we dissolve one mole of NaCl in water, one mole each of \[{\rm{N}}{{\rm{a}}^{\rm{ + }}}\] (sodium ions) and \[{\rm{C}}{{\rm{l}}^{\rm{-}}}\] (chloride ions) will be released into the solution.
If this occurs, then there would be two moles of ions in the solution.
One mole of NaCl after dissociation will form two moles in the solution.
So, \[{\rm{i = }}\frac{2}{1}\]
\[ \Rightarrow {\rm{i = 2}}\]

We know that
\[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = i}}{{\rm{k}}_{\rm{b}}}{\rm{m}}\]
where
\[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\]=elevation in boiling point
\[{{\rm{k}}_{\rm{b}}}\]=molal elevation constant
i=van't Hoff factor=1 as sucrose solution is an organic compound and won't dissociate in water.
So, \[\Delta {{\rm{T}}_{\rm{b}}}{\rm{ = 1 \times }}{{\rm{k}}_{\rm{b}}}{\rm{m = 0}}{\rm{.1^\circ C}}\]

For NaCl,
i=van't Hoff factor=2
\[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = 2 \times }}{{\rm{k}}_{\rm{b}}}{\rm{m}}\]
               \[ = 2 \times 0.1^\circ {\rm{C}}\]
               \[{\rm{ = 0}}{\rm{.2^\circ C}}\]

So, the elevation in boiling point for NaCl is \[{\rm{0}}{\rm{.2^\circ C}}\].

So, option B is correct.

Note: In the case of dissociation, the elevation in boiling point or depression in freezing point decreases. For example, ethanoic acid form dimer in benzene because of hydrogen bonding. This generally occurs in solvents possessing a low dielectric constant. So, the number of particles is decreased due to dimerization. Two molecules of ethanol acid after association form one molecule \[{\rm{i = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.5}}\]. Hence, as all the molecules of ethanoic acid collaborate in benzene, then elevation in boiling point or depression in freezing point decreases for ethanoic acid will be half of the normal value.