SATHEE: Solutions Topic Test Solutions Question 10

Question: Normal boiling point of water is 373 K (at 760 mm). Vapor pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mol, the boiling point of water at 23 mm pressure will be:

Options:

A) 250 K

B) 294 K

C) 51.6 K

D) 12.5 K

Answer:

Correct Answer: B

Solution:

[b] Applying ClausiusClapeyron equation, we get $\log \frac{P_{2}}{P_{1}} = \frac{Δ H_{v}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} \times T_{2}}]$ $\log \frac{760}{23} = \frac{40656}{2.303 \times 8.314} [\frac{373 - T_{1}}{373 T_{2}}]$ This gives $T_{1}$ = 294.4 K.

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Question: Normal boiling point of water is 373 K (at 760 mm). Vapor pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mol, the boiling point of water at 23 mm pressure will be:

Options:

Answer:

Solution:

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