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AIIMS 2009
Exam
AIIMS
Year
2009
Subject
Chemistry
Chapter
Solutions

AIIMS 2009 · Chemistry · Solutions

A 0.1 molal solution of an acid is 4.5 \% ionized. Calculate...

A 0.1 molal solution of an acid is 4.5%4.5 \% ionized. Calculate freezing point. (molecular weight of the acid is 300).Kf=1.86 K mol−1 kg) . K_f=1.86 \mathrm{~K} \mathrm{~mol}^{-1} \mathrm{~kg}.

Options

  1. A−0.194∘C-0.194^{\circ} \mathrm{C}
  2. B2.00∘C2.00^{\circ} \mathrm{C}
  3. C0∘C0^{\circ} \mathrm{C}
  4. D−0.269∘C-0.269^{\circ} \mathrm{C}

Correct answer

A. −0.194∘C-0.194^{\circ} \mathrm{C}

Step-by-step solution

If acid is 4.5%4.5 \% ionized then α=0.045\alpha=0.045. ΔTf= molality ×Kf=0.1×1.86=0.186ΔTexp⁡=ΔTf(1+α)=0.186(1+0.045)=0.194∘C\begin{aligned} & \Delta T_f=\text{ molality } \times K_f=0.1 \times 1.86=0.186 \\ & \Delta T_{\exp }=\Delta T_f(1+\alpha)=0.186(1+0.045)=0.194^{\circ} \mathrm{C}\end{aligned}
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