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Last updated date: 09th Dec 2023
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MVSAT Dec 2023

The boiling point of an aqueous solution of a non-volatile solute is \[100.15{}^\circ C.\]What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? The values of \[{{k}_{b}}\] ​and \[{{k}_{f}}\] for water are \[0.512\] and \[1.86K~~molalit{{y}^{-1}}\]
A.\[-0.544{}^\circ C.\]
B.\[-0.512{}^\circ C.\]
C.\[-0.272{}^\circ C.\]
D.\[-1.86{}^\circ C.\]

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Hint: We know that as we know that a non-volatile solute is a substance which does not readily evaporate into any gas under given conditions. It exhibits a low vapour pressure and high boiling point. Vapour pressure is the measure of change in the present state of a substance to its vapour state or gaseous state.

Complete answer:
We know that a non-volatile substance is the one which easily evaporates into gaseous form under given conditions of temperature and pressure. To calculate the vapour pressure we should have the mole fraction of the solute and solvent. Boiling point is reached when vapour pressure and atmospheric pressure becomes equal. We are given with the boiling point of water using which we can first calculate the boiling point elevation temperature using the formula:
As we know that; $\Delta {{T}_{b}}={{k}_{b}}m\Rightarrow 0.15=0.512\times m$
From here we get; $m=0.292;$
Similarly, $\Delta {{T}_{f}}={{k}_{f}}m'$ where $m'$ is $m'=\dfrac{m}{2}$ its predefined formulated value.
Now we also know that; \[{{T}_{1}}-{{T}_{0}}=1.86\times \dfrac{0.292}{2}=0.27{}^\circ C.\]
Thus from here we get value for ${{T}_{1}}$; ${{T}_{1}}=0-0.27=-0.27{}^\circ C.$
Therefore, the correct answer is Option C.
Also, the higher the vapour pressure of a substance, faster it will evaporate into vapours.

Remember that the vapour pressure is the tendency of a substance to change into the vapour phase or gaseous state from its original state which increases with increase in water and the boiling point is that temperature at which the vapour pressure of the substance at the surface of a liquid becomes equivalent to the pressure exerted by the surroundings.