Question:
Which of the following solution has the highest freezing point?
Which of the following solution has the highest freezing point?
Show Hint
Highest freezing point corresponds to the smallest value of:
\[
i\times m
\]
because freezing point depression is directly proportional to it.
Updated On: Jun 25, 2026
- \(0.1\ mol\ KCl\) in \(1\ kg\) water
- \(0.1\ mol\ K_2SO_4\) in \(1\ kg\) water
- \(0.1\ mol\) Urea in \(1\ kg\) water
- \(30\ g\) of glucose in \(1\ kg\) water
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use freezing point depression relation.
Freezing point depression is given by: \[ \Delta T_f=iK_fm \] where: \[ i=\text{van't Hoff factor} \] \[ m=\text{molality} \] Higher the value of: \[ i\times m \] greater is the depression in freezing point.
Hence, the solution with the smallest: \[ i\times m \] will have the highest freezing point.
Step 2: Calculate \(i\times m\) for each solution.
For \(KCl\): \[ KCl\rightarrow K^+ + Cl^- \] \[ i=2 \] Thus, \[ i\times m=2\times0.1=0.2 \] For \(K_2SO_4\): \[ K_2SO_4\rightarrow 2K^+ + SO_4^{2-} \] \[ i=3 \] Thus, \[ i\times m=3\times0.1=0.3 \] For urea:
Urea is a non-electrolyte: \[ i=1 \] Thus, \[ i\times m=1\times0.1=0.1 \] For glucose:
Molar mass of glucose: \[ 180\ g\ mol^{-1} \] Moles of glucose: \[ \frac{30}{180}=0.167\ mol \] Thus, \[ m=0.167 \] Since glucose is non-electrolyte: \[ i=1 \] Therefore, \[ i\times m=0.167 \]
Step 3: Compare the values.
\[ K_2SO_4 : 0.3 \] \[ KCl : 0.2 \] \[ \text{Glucose} : 0.167 \] \[ \text{Urea} : 0.1 \] Smallest value is for urea.
Hence, urea solution has minimum freezing point depression and therefore highest freezing point.
Step 4: Final conclusion.
Therefore, the solution with highest freezing point is \[ \boxed{0.1\ mol\ \text{Urea in}\ 1\ kg\ \text{water}} \]
Freezing point depression is given by: \[ \Delta T_f=iK_fm \] where: \[ i=\text{van't Hoff factor} \] \[ m=\text{molality} \] Higher the value of: \[ i\times m \] greater is the depression in freezing point.
Hence, the solution with the smallest: \[ i\times m \] will have the highest freezing point.
Step 2: Calculate \(i\times m\) for each solution.
For \(KCl\): \[ KCl\rightarrow K^+ + Cl^- \] \[ i=2 \] Thus, \[ i\times m=2\times0.1=0.2 \] For \(K_2SO_4\): \[ K_2SO_4\rightarrow 2K^+ + SO_4^{2-} \] \[ i=3 \] Thus, \[ i\times m=3\times0.1=0.3 \] For urea:
Urea is a non-electrolyte: \[ i=1 \] Thus, \[ i\times m=1\times0.1=0.1 \] For glucose:
Molar mass of glucose: \[ 180\ g\ mol^{-1} \] Moles of glucose: \[ \frac{30}{180}=0.167\ mol \] Thus, \[ m=0.167 \] Since glucose is non-electrolyte: \[ i=1 \] Therefore, \[ i\times m=0.167 \]
Step 3: Compare the values.
\[ K_2SO_4 : 0.3 \] \[ KCl : 0.2 \] \[ \text{Glucose} : 0.167 \] \[ \text{Urea} : 0.1 \] Smallest value is for urea.
Hence, urea solution has minimum freezing point depression and therefore highest freezing point.
Step 4: Final conclusion.
Therefore, the solution with highest freezing point is \[ \boxed{0.1\ mol\ \text{Urea in}\ 1\ kg\ \text{water}} \]
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