To calculate the heat of sublimation of iodine, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and enthalpy change:ln P2/P1 = -Hsub/R * 1/T2 - 1/T1 Where:P1 and P2 are the vapor pressures at temperatures T1 and T2, respectivelyHsub is the heat of sublimationR is the gas constant 8.314 J/molK We are given:P1 = 10 mmHg at T1 = -10C P2 = 760 mmHg at T2 = 184.35C, the normal boiling point First, we need to convert the temperatures to Kelvin:T1 = -10C + 273.15 = 263.15 KT2 = 184.35C + 273.15 = 457.5 KNow we can plug the values into the Clausius-Clapeyron equation:ln 760/10 = -Hsub/8.314 * 1/457.5 - 1/263.15 Solving for Hsub:Hsub = -8.314 * ln 76 / 1/457.5 - 1/263.15 Hsub 62083 J/molThe heat of sublimation of iodine is approximately 62,083 J/mol.