Question

Elevation in boiling point was $\text{0}\text{.52 }\\!\\!{}^\circ\\!\\!\text{ C}$ when 6 g of a compound X was dissolved in 100 g of water. The molecular mass of the compound M in $\text{gmo}{{\text{l}}^{-1}}$ is: (${{\text{K}}_{\text{b}}}$ for water is$\text{0}\text{.52 K}{{\text{m}}^{\text{-1}}}$)
(A) 120
(B) 60
(C) 600
(D) 180

Answer 1

The elevation of the boiling point $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}$ is a colligative property. It depends on the amount of solute. The difference in the boiling point is stated as:
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = }{{\text{K}}_{\text{b}}}\text{ m = }{{\text{K}}_{\text{b}}}\times \dfrac{{{\text{w}}_{\text{2}}}}{{{\text{M}}_{\text{2}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{w}}_{\text{1}}}}$
Where,
$\text{ }{{\text{K}}_{\text{b}}}\text{ }$is the ebullioscopic constant
$\text{ }{{\text{w}}_{\text{2}}}\text{ }$is the mass of solute
$\text{ }{{\text{w}}_{1}}\text{ }$is mass of solvent
$\text{ }{{\text{M}}_{\text{2}}}\text{ }$is the molar mass of solute

Complete step by step answer:
The boiling point, $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$ of a liquid, is the temperature at which the vapour pressure is equal to the atmospheric pressure. When a non-volatile solute is added to a liquid, the vapour pressure of the liquid is decreased. Hence, it must be heated to higher temperatures so that its vapour pressure becomes equal to that of the atmospheric pressure. This means that the addition of a non-volatile solute to a liquid raises its boiling point.
The elevation in boiling point $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$is related to the molar mass of the solute. The relation between the elevations in boiling point to the molality of solute is stated as follows:
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = }{{\text{K}}_{\text{b}}}\text{ m }$
If ${{\text{w}}_{\text{2}}}\text{ kg}$of the solute of the molar mass ${{\text{M}}_{\text{2}}}$ is dissolved in ${{\text{w}}_{1}}\text{ kg}$ of the solvent, then the number of moles of the solute dissolved in $\text{1 kg}$ of the solvent would be given by,
$\text{ m }=\text{ }\dfrac{{{\text{w}}_{\text{2}}}}{{{\text{M}}_{\text{2}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{w}}_{\text{1}}}}$
Then the equation becomes,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = }{{\text{K}}_{\text{b}}}\times \dfrac{{{\text{w}}_{\text{2}}}}{{{\text{M}}_{\text{2}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{w}}_{\text{1}}}}$ (1)
We are given the following data:
Elevation in boiling point, $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}=\text{ 0}\text{.52 }\\!\\!{}^\circ\\!\\!\text{ C}$
Weight of the compound X given,${{w}_{1}}\text{ = 6 g}$
Weight of solvent, ${{w}_{2\text{ }}}=\text{ 100 g}$
Ebullioscopic constant, ${{\text{K}}_{\text{b}}}\text{ = 0}\text{.52 K}{{\text{m}}^{-1}}$
We are interested to find the molecular weight of compound X.
On rearrangement of equation (1), we have,
$\begin{aligned} & {{\text{M}}_{\text{2}}}\text{ = }\dfrac{{{\text{K}}_{\text{b}}}\times \text{ }{{\text{w}}_{\text{2}}}}{\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{w}}_{\text{1}}}} \\\ & \text{0}\text{.52 = 0}\text{.52 }\times \text{ }\dfrac{\text{mass of solute}}{\text{molar mass of solute }\times \text{ mass of solvent in g}}\times 1000 \\\ \end{aligned}$

Let’s substitute the values in the equation of elevation in boiling point. We have,
${{\text{M}}_{\text{2}}}\text{ = }\left( 0.52\text{ K}{{\text{m}}^{\text{-1}}} \right)\times \dfrac{\text{6 g}}{\left( \text{0}\text{.5}{{\text{2}}^{\text{0}}}\text{C } \right)\text{ }\\!\\!\times\\!\\!\text{ 100 g}}\text{ }\times \text{1000 = 60 g mo}{{\text{l}}^{-1}}\text{ }$
Therefore, the molecular weight of compound X is equal to the $\text{60 g mo}{{\text{l}}^{-1}}\text{ }$.
Hence, (B) is the correct option.

Note: The elevation in boiling point can be related to the mole fraction of the solute. The relation is as shown below:
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = }\dfrac{\text{R T}_{\text{b}}^{\text{2}}}{{{\text{ }\\!\\!\Delta\\!\\!\text{ }}_{\text{vap}}}\text{H}}\text{ }\\!\\!\times\\!\\!\text{ }x\text{ }$
Where X is the mole reaction of solute and ${{\text{ }\\!\\!\Delta\\!\\!\text{ }}_{\text{vap}}}\text{H}$ is the heat of vaporization.
For dilute solutions, the mole fraction X is proportional to the molality thus we get the equation$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = }{{\text{K}}_{\text{b}}}\text{ m }$. Note that elevation in the boiling point is a colligative property; it depends on the amount of solute only. The concentration is always expressed in terms of molality.