Question

If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is _____ molal. (Rounded-off to the nearest integer) [\[{K_b} = 0.52\,{\rm{K}}\,{\rm{Kg}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\] ].

Answer 1

Hint: The elevation of boiling point defines the increase of boiling point of a solvent on the addition of a solute (non volatile) to it. Here, we will use the formula of b.p. elevation to get the molality of the solution.

Formula used: The formula for boiling point elevation is,
\[\Delta {T_b} = i{K_b}m\]
Here, \[\Delta {T_b}\] stands for boiling point elevation, i stands for Van’t Hoff factor, m is for molality, \[{K_b}\] is boiling point constant.

Complete Step by Step Answer:
Let’s find the value of i first.
AB undergoes dissociation to give two ions cation of A and anion of B. The dissociation of AB is as follows:
\[{\rm{AB}} \to {{\rm{A}}^ + } + {{\rm{B}}^ - }\]
The formula of i is,
\[i = 1 + (n - 1)\alpha \]
Here, n stands for number of ions and \[\alpha \] stands for percentage of dissociation. So, n=2 and \[\alpha = 75\% = \dfrac{{75}}{{100}} = 0.75\]
So,
\[i = 1 + (2 - 1)0.75 = 1 + 0.75 = 1.75\]
Now, all the values, \[\Delta {T_b}\], i, \[{K_b}\] are to be put in the equation of increase of boiling point.
Given, \[\Delta {T_b} = 2.5\,{\rm{K}}\]and \[{K_b} = 0.52\,{\rm{K}}\,{\rm{Kg}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\]
\[\Delta {T_b} = i{K_b}m\]
\[2.5 = 1.75 \times 0.52 \times m\]
\[m = \dfrac{{2.5}}{{1.75 \times 0.52}}\]
\[m = 2.75 \approx 3\]
Therefore, the molality value of the solution is 3 molal.

Note: Colligative property defines a solution’s property which depends on the ratio of the total molecules of solute to the total count of particles of solvent. Colligative properties do not depend on the chemical properties of the components of a solution. Some examples of the colligative properties are elevation of boiling point, depression of freezing point, lowering of vapour pressure etc.