Question
The latent heat of vaporization of water is 9800 $\mathrm{cal} / \mathrm{mole}$ and it the b.p. is $100^{\circ} \mathrm{C}$, the ebullioscopic constant of water isa. $0.511^{\circ}$b. $1.22^{\circ}$c. $51.18^{\circ}$d. $10.26^{\circ}$
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The latent heat of vaporization of water is 9700 cal/mol and if the boiling point is 100°C, the ebullioscopic constant of water is A. 0.513°Ckg/mol B. 1.026°Ckg/mol C. 10.26°Ckg/mol D. 1.832°Ckg/mol
A The heat of vaporization of water at $100^{\circ} \mathrm{C}$ is $2.26 \mathrm{~kJ} / \mathrm{g} ; \mathrm{at}$ $37^{\circ} \mathrm{C}$ (body temperature), it is $2.41 \mathrm{~kJ} / \mathrm{g}$. (a) Convert the latter value to standard molar heat of vaporization, $\Delta H_{\text {vap }}^{\circ}$, at \begin{array}{l} 37^{\circ} \mathrm{C} \text { . (b) Why is the heat of vaporization greater at } 37^{\circ} \mathrm{C}\\ \text { than at } 100^{\circ} \mathrm{C} \text { ? } \end{array}
For the vaporization of one mole of water at $100^{\circ} \mathrm{C}$ determine (a) $\Delta H$ (Table 8.3) (b) $\Delta P V$ (in kilojoules) (c) $\Delta E$
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