Question

A solution containing 0.2563 g of naphthalene (molecular mass = 128) in 50 g of carbon tetrachloride yields a boiling point elevation of $0.201{}^\circ C$ while a solution of 0.6216 g of an unknown solute in the same mass of the solvent gives a boiling point elevation of $0.647{}^\circ C$. Find the molecular mass of the unknown solute.

Answer 1

Hint: Try to recall that ${{K}_{b}}$ is called molal boiling point elevation constant and is a property of solvent. Think about the relation between the molal elevation constant, molality, and the change in temperature on the addition of a substance. Also, consider how we can calculate the molality of any given substance using the information given in the question.

Complete step by step solution:
-We know that elevation in the boiling point of a solution is calculated by $\Delta {{T}_{b}}={{K}_{b}}\times m$. Where, $\Delta {{T}_{b}}$ is the elevation in boiling point, ${{K}_{b}}$ is the molal elevation constant, and $m$ is the molality of the solution.
-In this problem, the value of ${{K}_{b}}$ is not given. So, first, we will calculate the ${{K}_{b}}$ which is further used to calculate the molar mass of the unknown solute.
-Calculation of ${{K}_{b}}$ for naphthalene:
Given,
weight of naphthalene = 0.2563 g
The molecular mass of naphthalene=128
Elevation in boiling point of naphthalene solution, $\Delta {{T}_{b}}=0.201{}^\circ C$
Weight of solvent, $W=50g=0.05kg$
The two formulae that we will use to calculate the molality are, the actual formula for molality and the formula to calculate the number of moles of any substance. They are as follows:
\[\begin{align}
& \text{molality}=\dfrac{\text{number of moles of solute}}{\text{mass of solvent in kg}} \\
& \text{number of moles}=\dfrac{\text{given weight of solute}}{\text{molecular weight of solute}} \\
\end{align}\]
First, we will calculate the number of moles of the substance (naphthalene) present using the second formula and then the molality. So, solving for the number of moles, we get:
\[\begin{align}
& \text{number of moles}=\dfrac{0.2563}{128} \\
& \text{number of moles}=0.002 \\
\end{align}\]
Now, that we have the number of moles of naphthalene, we can find its molality using the given formula. Plugging in all the values and solving for molality, we get:
\[\begin{align}
& \text{molality}=\dfrac{0.002}{0.05} \\
& m=0.04 \\
\end{align}\]
Now, to find the molal elevation constant for the solvent carbon tetrachloride, we will put the values that we have obtained in the equation. So, we will get:
\[\begin{align}
& \Delta {{T}_{b}}={{K}_{b}}\times m \\
& 0.201={{K}_{b}}\times (0.04) \\
& {{K}_{b}}=5.025K\text{ }kgmo{{l}^{-1}} \\
\end{align}\]
Now, we move on to finding the molecular mass of the unknown solute. For this we have to know the molality of the solution. The molality can be obtained by the same equation that relates the change in boiling point, the molal elevation constant, and the molality since we have the values of $\Delta {{T}_{b}}$ and ${{K}_{b}}$ now.
-So, putting these values in the equation, and considering
$m'$ as the molality of the unknown solvent
$M$ as the molecular mass of the unknown solute
Weight of unknown solute, $w'=0.6216g$
Elevation in boiling point of unknown solute, $\Delta {{T}_{b}}=0.647{}^\circ C$
Weight of solvent, $W=50g=0.05kg$
So, the molality of the solute will be:
\[\begin{align}
& m'=\dfrac{\Delta {{T}_{b}}}{{{K}_{b}}} \\
& m'=\dfrac{0.647}{5.025} \\
& m'=0.129molal \\
\end{align}\]
Since we now have the molality, we can find the number of moles of the unknown solute. Putting the values in the equation, we get:
\[\begin{align}
& m'=\dfrac{n'}{W} \\
& n'=0.129\times 0.05 \\
& n'=0.00645 \\
\end{align}\]
Now, calculate the molecular mass of the unknown solute using the equation for the number of moles of solute. Putting in the values, we get:
\[\begin{align}
& n'=\dfrac{w'}{M} \\
& M=\dfrac{w'}{n'} \\
& M=\dfrac{0.6216}{0.00645} \\
& M=96.37g/mol \\
\end{align}\]

Hence, the molecular mass of the unknown solute is 96.37 g/mol.

Note: Note that elevation in boiling point is a colligative property, but the boiling point is not a colligative property. Also, you should remember that the boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure. While calculating the molality, be careful about the value in the denominator, other terms for concentration usually have the characteristics of the solution in the denominator like the volume or weight of the solution; but in molality, we only consider the weight of the solvent and not the solution.