The boiling point of toluene is 110.7,^oC and | vedclass

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(A) The molal elevation constant $(K_b)$ is related to the enthalpy of vaporization $(\Delta_{vap}H)$ by the formula: $K_b = \frac{R 	imes T_b^2 	im

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$34$

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(A) The molal elevation constant $(K_b)$ is related to the enthalpy of vaporization $(\Delta_{vap}H)$ by the formula: $K_b = \frac{R 	imes T_b^2 	imes M}{1000 	imes \Delta_{vap}H}$.Here,$T_b = 110.7   273.15 = 383.85\, K$.$R = 8.314\, J\, K^{-1}\, mol^{-1}$.$M$ (molar mass of toluene,$C_7H_8$) = $92\, g\, mol^{-1}$.Rearranging for $\Delta_{vap}H$: $\Delta_{vap}H = \frac{R 	imes T_b^2 	imes M}{1000 	imes K_b}$.$\Delta_{vap}H = \frac{8.314 	imes (383.85)^2 	imes 92}{1000 	imes 3.32}$.$\Delta_{vap}H = \frac{8.314 	imes 147340.82 	imes 92}{3320} \approx 33945\, J\, mol^{-1} \approx 34\, kJ\, mol^{-1}$.

The boiling point of toluene is $110.7\,^oC$ and its molal elevation constant is $3.32\, K\, kg\, mol^{-1}$. The enthalpy of vaporization of liquid toluene is ............ $kJ\, mol^{-1}$.

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