NEET Chemistry NEET PYQ-Solutions

  • question_answer
    A 0.1 molal aqueous solution of a weak acid is 30% ionised. If \[{{K}_{f}}\] for water is \[1.86{}^\circ C/m,\] the freezing point of the solution will be [AIPMT (M) 2011]

    A) \[\text{ }0.18{}^\circ C\]

    B) \[\text{ }0.54{}^\circ C~\]

    C) \[\text{ }0.36{}^\circ C\]

    D) \[\text{ }0.24{}^\circ C\]

    Correct Answer: D

    Solution :

    [d] Freezing point depression\[(\Delta {{T}_{f}})=i{{K}_{f}}m\]
    \[\underset{\begin{smallmatrix}  1-\alpha  \\  1-0.3 \end{smallmatrix}}{\mathop{\text{HA}}}\,\xrightarrow[{}]{{}}\underset{\begin{smallmatrix}  \alpha  \\  0.3 \end{smallmatrix}}{\mathop{{{\text{H}}^{\text{+}}}}}\,+\underset{\begin{smallmatrix}  \alpha  \\  0.3 \end{smallmatrix}}{\mathop{{{\text{A}}^{-}}}}\,\]
    \[i=1-0.3+0.3+0.3\]
    \[i=1.3\]
    \[\therefore \]      \[\Delta {{T}_{f}}=1.3\times 1.86\times 0.1={{0.2418}^{o}}C\]
                \[{{T}_{f}}=0-{{0.2418}^{o}}C\]
                \[=-{{0.2418}^{o}}C\]


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