Question

Solution containing \[0.73{\text{ }}g\] camphor (Molar mass $152gmo{l^{( - 1)}}$) in \[36.8{\text{ }}g\]of acetone (boiling point \[{56.3^ \circ }\] C boil at ${56.55^ \circ }$ C. A solution of \[0.564{\text{ }}g\] unknown compound in same weight of acetone boils at ${56.46^ \circ }$ C. Calculate the molar mass of the unknown compound.

Answer 1

Hint: As we all know about colligative properties and the method to determine Molar Mass then we will use that concept here. The first thing is that we need to keep in mind about the molality of a solution. Also, let’s keep in mind that the colligative properties include lowering of vapor pressure, the elevation of boiling point, depression of freezing point, and osmotic pressure.

Complete step by step answer:
Now, let’s calculate the elevation constant ${K_b}$ for camphor:
For that, we need to see all the given contents.
Mass of a solvent, ${W_A} = {\text{ }}36.8{\text{ }}g$
Mass of a solute, ${W_B} = {\text{ }}0.73{\text{ }}g$
And the molecular mass of a solute, ${M_b} = {\text{ }}152$
As we all know that the boiling point of an elevation is $\vartriangle {T_b}$ then let’s calculate the value of an elevation.
$\Delta {T_b}\; = {\text{ }}56.55{\text{ }}-{\text{ }}56.30$
$\Delta {T_b} = 0.25^\circ C$
Now, to calculate the value of ${K_b}$ we need to consider the formula of total elevation. So, let’s remind the formula first.
The formula of an elevation,
$\vartriangle {T_b} = {K_b} \times \dfrac{{{W_B} \times 1000}}{{{W_A} \times {M_b}}}$
Now, we’ll place the values of the respective abbreviations and then we will solve the formula to find the value of ${K_b}$
${K_B} = \dfrac{{\Delta {T_B} \times {W_A} \times {M_b}}}{{{W_B} \times 1000}}$
Now, we have the value of ${K_b}$so we will determine the value of it.
${K_b}$= $\dfrac{{0.25 \times 36.8 \times 152}}{{0.73 \times 1000}}$
${K_b}$= ${1.9156^ \circ }kgmo{l^{ - 1}}$
After calculating the value of ${K_b}$, we will refer to the other part of the question where we need to calculate the molecular mass of an unknown solute.
Again, we will note down all the given values and follow the same steps to calculate the value of ${M_b}$ i.e the mass of the unknown solute.
Mass of a solvent, ${W_A} = 36.8g$
Mass of a solute, ${W_B} = 0.564g$
So, again we find elevation in boiling point
$
  \Delta {T_b} = 56.46 - 56.30 \\
  \Delta {T_b} = {0.16^ \circ }C \\
 $
Now, we will write the formula in terms of ${M_b}$ .
$\vartriangle {T_b} = {K_b} \times \dfrac{{{W_B} \times 1000}}{{{W_A} \times {M_B}}}$
${M_b} = \dfrac{{{K_b} \times {W_B} \times 1000}}{{{W_A} \times \Delta {T_b}}}$
Now, we will place the values of the given abbreviations to calculate the mass of unknown solute in grams.
$
  {M_b} = \dfrac{{1.9156 \times 0.564 \times 1000}}{{36.8 \times 0.16}} \\
  {M_b} = 183.4g \\
 $
So, we have got the mass of an unknown solute as 183.4 g.

Note:
We must know that the camphor is terpene (organic compound) commonly utilized in creams, ointments, and lotions. Camphor is the oil extracted from the wood of camphor trees and processed by steam distillation. It is often used topically to alleviate pain, irritation, and itching. Camphor is additionally wont to relieve chest congestion and inflammatory conditions.