A) \[-{{4.80}^{o}}C\]
B) \[-{{0.360}^{o}}C\]
C) \[-{{0.260}^{o}}C\]
D) \[-{{2.56}^{o}}C\]
Correct Answer: A
Solution :
\[\begin{matrix} HX \\ 1 \\ 1-0.3 \\ \end{matrix}\begin{matrix} \\ {} \\ {} \\ \end{matrix}\,\,\begin{matrix} {{H}^{+}} \\ 0 \\ 0.3 \\ \end{matrix}\,\,\begin{matrix} + \\ {} \\ {} \\ \end{matrix}\,\,\,\begin{matrix} {{X}^{-}} \\ 0 \\ 0.3 \\ \end{matrix}\] Total number of moles after dissociation \[=1-0.3+0.3+0.3=1.3\] \[\frac{{{k}_{f}}observed}{{{k}_{f}}\exp erimental}=\frac{No.\text{ }of\text{ }moles\text{ }after\text{ }dissociation}{No.\text{ }of\text{ }moles\text{ }before\text{ }dissociation}\]\[\frac{{{k}_{f}}\,observed}{1.85}=\frac{1.3}{1}\] \[{{k}_{f}}\] observed \[=1.85\times 1.3=2.405\] \[\Delta {{T}_{f}}={{k}_{f}}\times \] molality \[\Delta {{T}_{f}}=2.405\times 0.2={{0.481}^{o}}C\] \[\therefore \] Freezing point of solution \[=0-0.481=-{{0.481}^{o}}C\]You need to login to perform this action.
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