Calculate the total quantity of heat required to convert 25.0 g of liquid CCl₄(l)from 35.0°C to gaseous CCl₄ at 76.8°C (the normal boiling point for CCl₄).The specific heat of CCl₄(l)is 0.857 J/(g ∙ °C),its heat of fusion is 3.27 kJ/mol,and its heat of vaporization is 29.82` kJ/mol.

Question

Quizplus · Accepted Answer

The total heat required can be calculated as the sum of the heat required to heat the liquid from 35.0°C to its boiling point, the heat required to vaporize the liquid, and the heat required to heat the vapor from its boiling point to 76.8°C. Heat required to heat the liquid: q = m × c × ΔT q = 25.0 g × 0.857 J/(g ∙ °C) × (76.8°C - 35.0°C) q = 927.35 J Heat required to vaporize the liquid: q = n × ΔHvap First, we need to calculate the number of moles of CCl₄(l) present. n = m/M n = 25.0 g / 153.81 g/mol n = 0.1624 mol Now we can calculate the heat required to vaporize the liquid. q = 0.1624 mol × 29.82 kJ/mol q = 4.832 kJ Heat required to heat the vapor: q = m × c × ΔT q = 25.0 g × 0.857 J/(g ∙ °C) × (76.8°C - 76.8°C) q = 0 J (no temperature change required) Total heat required: qtotal = 927.35 J + 4.832 kJ + 0 J qtotal = 5.759 kJ ≈ 5.74 kJ (rounded to three significant figures) Therefore, the best choice is C, which matches the calculated value of 5.74 kJ.

Calculate the Total Quantity of Heat Required to Convert 25