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# 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.$-1.86{}^\circ C.$

Last updated date: 22nd Jul 2024
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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.

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 ${{T}_{1}}$; ${{T}_{1}}=0-0.27=-0.27{}^\circ C.$