Question

For a dilute solution containing a non-volatile solute, the molar mass of solute evaluated from the elevation of boiling point is given by the expression:
A. ${M_2} = \dfrac{{\Delta {T_b}}}{{{k_b}}}\dfrac{{{m_1}}}{{{m_2}}}$
B. ${M_2} = \dfrac{{\Delta {T_b}}}{{{k_b}}}\dfrac{{{m_2}}}{{{m_1}}}$
C. ${M_2} = \dfrac{{{k_b}}}{{\Delta {T_b}}}\dfrac{{{m_2}}}{{{m_1}}}$
D. ${M_2} = \dfrac{{{k_b}}}{{\Delta {T_b}}}\dfrac{{{m_1}}}{{{m_2}}}$

Answer 1

We can use the relationship between the molality of the solution and elevation of boiling point along with that of amount and molar mass.

Complete step by step solution
Among other colligative properties, we have elevation of boiling point of a solvent that also results from the lowering in vapor pressure of the solvent upon addition of a non-volatile solute.

The elevation of boiling point can be related to the molality of the solution by the following expression:
$\Delta {T_b} = {K_b}m$
Here, $\Delta {T_b}$ is the elevation of boiling point of the solvent, ${K_b}$ is the molal
elevation constant for the given solvent and is the molality of the solution.

We can also write the molality of a solution by using number of moles of solute $\left( {{n_{solute}}} \right)$ present in the given mass of solvent $\left( {{m_{solvent}}} \right)$ as follows:
$m = \dfrac{{{n_{solute}}}}{{{m_{solvent}}\;in\;kg}}$

We know that the mass of $1$ mole of a substance is its molar mass $M$ . So we can write:
$\dfrac{{1\;{\rm{mol}}}}{M}$
We can use this conversion factor to calculate the amount of solute in given mass $\left( {{m_{solute}}} \right)$ by using its molar mass $\left( {{M_{solute}}} \right)$as follows:

$${n_{solute}} = {m_{solute}} \times \left( {\dfrac{{1\;{\rm{mol}}}}{{{M_{solute}}\;g}}} \right)\\\ = \dfrac{{{m_{solute}}}}{{{M_{solute}}}}$$

Let’s rewrite the previously written expression of molality by using the above expression for amount of solute as follows:
$m = \dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}$

We can use this equation to rewrite the expression for elevation of boiling point as follows:
\Delta {T_b} = {K_b}\left\\{ {\dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}} \right\\}
Let’s rearrange this equation for molar mass of solute:
{K_b}\left\\{ {\dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}} \right\\} = \Delta {T_b}\\\ \dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}} = \dfrac{{\Delta {T_b}}}{{{K_b}}}\\\ {M_{solute}} = \dfrac{{{K_b}}}{{\Delta {T_b}}}\dfrac{{{m_{solute}}}}{{{m_{solvent}}\;in\;kg}}

Here, we have used subscript $1$ for solvent and $2$ for solute. So, let’s incorporate these in our derived equation as follows:
${M_2} = \dfrac{{{K_b}}}{{\Delta {T_b}}}\dfrac{{{m_2}}}{{{m_1}}}$

**Hence, the correct option is C.

Note:**
We have to use the subscripts carefully for solute and solvent in the expressions.
We also have to take care of the units as well.