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TS EAMCET · Chemistry · Solutions

An aqueous solution of a non-volatile solute boils at \(100.17^{\circ} \mathrm{C}\). The temperature at which this solution will freeze \(\left(\right.\) in \(\left.{ }^{\circ} \mathrm{C}\right)\) is \(\begin{aligned} & \left(\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.512^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1},\right. \ & \left.\qquad \mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\right)\end{aligned}\)

  1. A \(-0.62\)
  2. B \(-0.512\)
  3. C \(-1.24\)
  4. D \(-1.86\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-0.62\)

Step-by-step Solution

Detailed explanation

Using formula, \(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{k}_{\mathrm{b}}\) molality \((100.17-100)^{\circ} \mathrm{C}=0.512^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1} \times\) molality molality \(=\frac{0.17}{0.512} \mathrm{~m}\) Now, depression in freezing point is given…