AP EAMCET · Chemistry · Solutions
0.05 mole of a non-volatile solute is dissolved in \(500 \mathrm{~g}\) of water. What is the depression in freezing point of resultant solution?
\(\left(K_f\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\)
- A \(0.047 \mathrm{~K}\)
- B \(0.372 \mathrm{~K}\)
- C \(0.093 \mathrm{~K}\)
- D \(0.186 \mathrm{~K}\)
Answer & Solution
Correct Answer
(D) \(0.186 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
Given, \(n=0.05 \mathrm{~mol}\) \(\begin{aligned} V & =500 \mathrm{~g} \text { of water } \\ K_f & =1.86 \mathrm{k} \mathrm{kg} / \mathrm{mol} \\ \Delta T_f & =?\end{aligned}\) \(\therefore\) Molality…
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