QJEEM02Ch948d: JEE Main 2002 Chemistry - Class 12 JEEM02Ch948d Solution: Freezing point of an aqueous solution is (-0.186) o C.…

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QJEEM02Ch948d: JEE Main 2002 Chemistry - Class 12 JEEM02Ch948d Solution: Freezing point of an aqueous solution is (-0.186) o C.…

Freezing point of an aqueous solution is (-0.186)^oC. Elevation of boiling point of the same solution is K_b = 0.512 ^oC, K_f = 1.86 ^oC, find the increase in boiling point.

Multiple Choice Options

A 0.186 oC
B 0.0512 oC
C 0.092 oC
D 0.2732 oC

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Accepted Answer

Δ T_{b} = K_{b} \frac{W_{B}}{M_{B} \times W_{A}} \times 1000;

Δ T_{f} = K_{f} \frac{W_{B}}{M_{B} \times W_{A}} \times 1000;

\frac{Δ T_{b}}{Δ T_{f}} = \frac{K_{b}}{K_{f}} = \frac{Δ T_{b}}{- 0.186}

= \frac{0.512}{1.86}

= {0.0512}^{\circ} C .

QJEEM02Ch948d: JEE Main 2002 Chemistry - Class 12 JEEM02Ch948d Solution: Freezing point of an aqueous solution is (-0.186) o C.…

Multiple Choice Options

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