Question

Acetone boils ${56.38^{\,\,o}}C$ at and a solution of $1.41$ grams of an organic compound in $20g$ of acetone boils at ${56.88^{\,\,o}}C$. If ${K_b}$for acetone is $1.67\,K\,Kg\,mo{l^{ - 1}}$. Calculate the mass of one mole of the organic solid.

Answer 1

In this question, when an organic compound is added in acetone, there is elevation in the boiling point of acetone. Acetone has a chemical formula of $C{H_3}COC{H_3}$ and the molecular weight of acetone $58.08\,g/mol$.
Formula used- $\Delta {T_b} = \,{K_b}\,\, \times \,\,\dfrac{{Weight\,of\,solute\,({W_B})}}{{Mass\,of\,solute\,({M_B})}}\,\, \times \dfrac{{1000}}{{Mass\,of\,solvent\,({M_A})}}$

Complete answer:
As it is given in the question, when $1.41$ gram of organic compound is added in acetone. The boiling point of acetone gets elevated. This question is based on the colligative property of solution i.e. Elevation in boiling point
Elevation in boiling point- this colligative property describes the phenomenon that when another compound i.e. solute is added to the liquid (or solvent), the boiling point of the solvent will be higher. In other words, the solution i.e. mixture of solute and solvent has a higher boiling point than a pure solvent.
This elevation in the boiling point is denoted by $\Delta {T_b}$ which is directly proportional to the molality of the solution$(m)$.
$\Delta {T_b} \propto \,m$
Removing this proportionality, a constant is added.
We get, $\Delta {T_b} = \,{K_b}\,\,m$
Where, ${K_b}\,$= ebullioscopic constant
We know, molality $(m)\, = \,\dfrac{{Weight\,of\,solute\,({W_B})\,in\,grams}}{{Molar\,Mass\,of\,solute\,({M_B})}}\,\, \times \dfrac{{1000}}{{Mass\,of\,solvent\,({M_A})\,in\,grams}}$
By substituting the formula of molality, we get,
$\Delta {T_b} = \,{K_b}\,\, \times \,\,\dfrac{{Weight\,of\,solute\,({W_B})\,in\,grams}}{{Molar\,Mass\,of\,solute\,({M_B})}}\,\, \times \dfrac{{1000}}{{Mass\,of\,solvent\,({M_A})\,in\,grams}}$
The above question is based on this formula.
Given:
Boiling point of pure acetone ${T_b}^o\, = \,{56.38^{\,\,o}}C$
Boiling point of the solution ${T_b}\, = \,{56.88^{\,\,o}}C$
So, $\Delta {T_b}\,\, = \,\,{T_{b\,\,}}\, - \,\,{T_b}^o\,$
Weight of solute ${W_B}\, = 1.41\,g$
Mass of solvent ${M_A}\, = 20\,g$
${K_b}\, = \,1.67\,K\,Kg\,mo{l^{ - 1}}$
We have to calculate molar mass of solute $({M_B})\,$
Using this formula,
$\Delta {T_b} = \,{K_b}\,\, \times \,\,\dfrac{{Weight\,of\,solute\,({W_B})\,in\,grams}}{{Molar\,Mass\,of\,solute\,({M_B})}}\,\, \times \dfrac{{1000}}{{Mass\,of\,solvent\,({M_A})\,in\,grams}}$

$\Delta {T_b}\,\, = \,\,{T_{b\,\,}}\, - \,\,{T_b}^o\, = \,{K_b}\,\, \times \,\,\dfrac{{Weight\,of\,solute\,({W_B})\,in\,grams}}{{Molar\,Mass\,of\,solute\,({M_B})}}\,\, \times \dfrac{{1000}}{{Mass\,of\,solvent\,({M_A})\,in\,grams}}$
Substituting all the values we get,
$56.88 - 56.38\, = \,1.67\,\, \times \,\,\dfrac{{1.41}}{{Molar\,Mass\,of\,solute\,({M_B})}}\,\, \times \dfrac{{1000}}{{20}}$
$0.5\, = \,1.67\,\, \times \,\,\dfrac{{1.41}}{{Molar\,Mass\,of\,solute\,({M_B})}}\,\, \times \dfrac{{1000}}{{20}}$

After solving this, we get
Molar mass of solute $\,({M_B}) = \,\dfrac{{1.67\, \times \,1.41\, \times 1000}}{{0.5\, \times 20}}$
Molar mass of solute $({M_B})\, = \,235.47\,g/mol$
Hence, the mass of one mole of the organic solid $({M_B})\, = \,235.47\,g/mol$

Note:
The elevation in boiling point can be explained in terms of vapor pressure. A liquid boils at the temperature when vapor pressure is equal to surrounding pressure. When a nonvolatile solute is added, it has zero vapor pressure. So, the vapor pressure of the solution is less than the solvent which indicates that higher temperature is needed for the vapor pressure to reach surrounding pressure and hence, the boiling point is elevated.