Question

Solution-1 : \(2.025\,\text{g}\) glucose, \(125\,\text{mL}\)
Solution-2 : \(9\,\text{g}\) urea, \(500\,\text{mL}\)
Solution-3 : \(1.9\,\text{g}\) CaCl$_2$, \(250\,\text{mL}\)
Solution-4 : \(20.5\,\text{g}\) Al$_2$(SO$_4$)$_3$, \(750\,\text{mL}\)
[2mm] Order of \(\Delta T_b\) is:

• A. Al$_2$(SO$_4$)$_3$ > Urea > CaCl$_2$ > Glucose
• B. Al$_2$(SO$_4$)$_3$ > CaCl$_2$ > Urea > Glucose
• C. Glucose > Al$_2$(SO$_4$)$_3$ > CaCl$_2$ > Urea
• D. CaCl$_2$ > Urea > Glucose > Al$_2$(SO$_4$)$_3$

Hint:

For colligative properties, always compare \(\boldsymbol{i \times m}\). Electrolytes dominate due to higher van’t Hoff factor.

Answer 1

The Correct Option is A

The problem asks us to compare the elevation in boiling point (ΔT_b) for different solutions.

The formula for boiling point elevation is:

ΔT_b = i × K_b × m

where:

• i = van’t Hoff factor
• K_b = ebullioscopic constant (same solvent, hence constant)
• m = molality

Since K_b is the same for all solutions, ΔT_b depends on the product i × m.

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Solution-wise analysis:

1. Glucose (C₆H₁₂O₆)
   Glucose does not dissociate, so i = 1.
   Molality ≈ 0.112 mol/kg
   i × m = 0.112
2. Urea (CO(NH₂)₂)
   Urea does not dissociate, so i = 1.
   Molality ≈ 0.30 mol/kg
   i × m = 0.30
3. Calcium chloride (CaCl₂)
   CaCl₂ dissociates into 3 ions.
   i = 3
   Molality ≈ 0.0685 mol/kg
   i × m = 0.2055
4. Aluminium sulfate (Al₂(SO₄)₃)
   Al₂(SO₄)₃ dissociates into 5 ions.
   i = 5
   Molality ≈ 0.08 mol/kg
   i × m = 0.40

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Comparison of ΔT_b values:

Al₂(SO₄)₃ > Urea > CaCl₂ > Glucose

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Final Answer:

Al₂(SO₄)₃ > Urea > CaCl₂ > Glucose