The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\). The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(g)\) are \(0.91 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K},\) respectively. The heat of vaporization for the compound is \(27.49 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the heat required to convert \(35.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) from a liquid at \(10.00^{\circ} \mathrm{C}\) to a gas at \(105.00^{\circ} \mathrm{C}\).

To calculate the heat required to heat the liquid from its initial temperature to boiling point, we will use the formula: \(q = mc\Delta T\), where \(q\) is the heat, \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. Given data: \(m = 35.0 \mathrm{g}\), \(c = 0.91 \mathrm{ J / gK} \), \(\Delta T = T_\mathrm{final} - T_\mathrm{initial} = T_\mathrm{boil} - T_\mathrm{initial} = 47.6^{\circ}\mathrm{C} - 10.00^{\circ}\mathrm{C} = 37.6^{\circ}\mathrm{C}\), We can now calculate the heat required: \(q_1 = (35.0 \mathrm{g}) (0.91 \mathrm{ J / gK}) (37.6\mathrm{K}) = 1209.71 \mathrm{J}\).

To calculate the heat required to vaporize the liquid, we will use the formula: \(q = n \Delta H_\mathrm{vap}\), where \(n\) is the number of moles, and \(\Delta H_\mathrm{vap}\) is the heat of vaporization. Given data: \(\Delta H_\mathrm{vap} = 27.49 \mathrm{ kJ/mol} \), First, we need to calculate the molar mass of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}\): Molar mass = (2 × 12.01) + (3 × 35.45) + (3 × 19.00) = 24.02 + 106.35 + 57.00 = 187.37 \(\mathrm{g/mol}\). Now we calculate the number of moles: \(n = \frac{m}{M} = \frac{35.0\mathrm{g}}{187.37\mathrm{g/mol}} = 0.1869\mathrm{mol}\), Finally, we can calculate the heat required to vaporize the liquid: \(q_2 = (0.1869\mathrm{mol}) (27.49 \times 10^3 \mathrm{J/mol}) = 5139.97\mathrm{J}\).

To calculate the heat required to heat the gas from its boiling point to the final temperature, we will use the formula: \(q = mc\Delta T\), where \(q\) is the heat, \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. Given data: \(m = 35.0 \mathrm{g}\), \(c = 0.67 \mathrm{ J / gK} \), \(\Delta T = T_\mathrm{final} - T_\mathrm{initial} = T_\mathrm{final} - T_\mathrm{boil} = 105.00^{\circ}\mathrm{C} - 47.6^{\circ}\mathrm{C} = 57.4^{\circ}\mathrm{C}\), We can now calculate the heat required: \(q_3 = (35.0 \mathrm{g}) (0.67 \mathrm{ J / gK}) (57.4\mathrm{K}) = 1352.21 \mathrm{J}\).

To find the total heat required to convert the compound from liquid at \(10.00^{\circ} \mathrm{C}\) to gas at \(105.00^{\circ} \mathrm{C}\), we just add the heat from the previous steps: Total heat (\(q_\mathrm{total}\)) = \(q_1 + q_2 + q_3 = 1209.71\mathrm{J} + 5139.97\mathrm{J} + 1352.21\mathrm{J} = 7702.89\mathrm{J}\). Therefore, the heat required to convert \(35.0 \mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}\) from a liquid at \(10.00^{\circ} \mathrm{C}\) to a gas at \(105.00^{\circ} \mathrm{C}\) is approximately \(7702.89 \mathrm{J}\).