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AP EAMCET · Chemistry · Solutions

\(1.8 \mathrm{~g}\) of glucose (molar mass \(180 \mathrm{~g} \mathrm{~mol}^{-1}\) ) is dissolved in \(0.1 \mathrm{~kg}\) of water. The freezing point of the solution ( \(\mathrm{in}^{\circ} \mathrm{C}\) ) is
\(\left(K_f\right.\) for water \(\left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\)

  1. A \(+0.186\)
  2. B \(-0.372\)
  3. C \(-0.186\)
  4. D \(+0.372\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(-0.186\)

Step-by-step Solution

Detailed explanation

Depression in freezing point is given as \(\Delta T=i K_f \times m\) where, \(i=\) van't Hoff factor \(\begin{gathered}K_f=\text { molal freezing point depression } \\ \text { constant }\end{gathered}\) \(m=\) molality Molar mass of glucose…