MHT CET · Chemistry · Solutions
What is boiling point of a decimolal aqueous solution of glucose if molal elevation constant for water is \(0.52^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\) ?
- A \(101.52^{\circ} \mathrm{C}\)
- B \(99.95^{\circ} \mathrm{C}\)
- C \(99.48^{\circ} \mathrm{C}\)
- D \(100.052^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(100.052^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{k}_{\mathrm{b}} \cdot \mathrm{m} \)
\( \Delta \mathrm{T}_{\mathrm{b}}=0.52 \times 0.1, \mathrm{~m}=0.1 \text { (decimolal solution) } \)
\( =0.052^{\circ} \mathrm{C} \)
\( \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^{\circ} \quad\left(\mathrm{T}_{\mathrm{b}}^{\circ}=\right.\text { Boiling point of}\) \(\text{pure water})\)
\( 0.052^{\circ} \mathrm{C}=\mathrm{T}_{\mathrm{b}}-100^{\circ} \mathrm{C} \)
\( \mathrm{T}_{\mathrm{b}}=100.052^{\circ} \mathrm{C}\)
\( \Delta \mathrm{T}_{\mathrm{b}}=0.52 \times 0.1, \mathrm{~m}=0.1 \text { (decimolal solution) } \)
\( =0.052^{\circ} \mathrm{C} \)
\( \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^{\circ} \quad\left(\mathrm{T}_{\mathrm{b}}^{\circ}=\right.\text { Boiling point of}\) \(\text{pure water})\)
\( 0.052^{\circ} \mathrm{C}=\mathrm{T}_{\mathrm{b}}-100^{\circ} \mathrm{C} \)
\( \mathrm{T}_{\mathrm{b}}=100.052^{\circ} \mathrm{C}\)
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