SolutionHard
Question
The molar latent heat of vaporization of a certain liquid is 4.9 kcal/mol. The normal boiling point of the liquid is 77°C. The change in boiling point per atm increase in pressure at the normal boiling point of the liquid is
Options
A.50 deg/atm
B.0.02 deg/atm
C.5.0 deg/atm
D.0.05 deg/atm
Solution
Clausius–Clapeyron equation:
$\frac{dT}{dP} = \frac{RT^{2}}{\Delta H.P} = \frac{2 \times (350)^{2}}{4.9 \times 10^{3}} = 50\text{ K/atm}$
Create a free account to view solution
View Solution FreeMore Solution Questions
pH of 0.02 M aq. solution of NH4Cl (Pkb = 4.73) is :-...The vapour pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373 K. The mole fraction of solute i...The molecular mass of glucose is 180 g/ mol and that of sucrose is 342 g/mol. Assuming ideal behaviour, a plot of freezi...Van′t Hoff factors of aqueous solutions of X, Y, Z are 1.8, 0.8 and 2.5. Hence, their (assume equal concentrations...Which combination of (I) vapour pressure, (II) intermolecular forces and (III) ᐃHvap (latent heat of vaporisation)...