Question

$\text{Standard enthalpy of vaporisation for CCl}_4 \text{ is 30.5 kJ mol}^{-1}. \text{ Heat required for vaporisation of } 284 \, \text{g of CCl}_4 \text{ at constant temperature is } \_\_\_\_ \, \text{kJ.} \\ \text{(Given molar mass in g mol}^{-1}\text{; C = 12, Cl = 35.5)}$

Answer 1

Correct Answer: 56

The standard enthalpy of vaporisation \((\Delta H_v)\) for \(\text{CCl}_4\) is provided as \(30.5 \, \text{kJ mol}^{-1}\). The objective is to determine the heat necessary to vaporize \(284 \, \text{g}\) of \(\text{CCl}_4\). First, calculate the molar mass of \(\text{CCl}_4\). The molecular formula is \(\text{CCl}_4\), comprising 1 carbon (C) atom and 4 chlorine (Cl) atoms:

• \(\text{Molar mass of C} = 12 \, \text{g mol}^{-1}\)
• \(\text{Molar mass of Cl} = 35.5 \, \text{g mol}^{-1}\)
• Molar mass of \(\text{CCl}_4 = (1 \times 12) + (4 \times 35.5) = 12 + 142 = 154 \, \text{g mol}^{-1}\)

Subsequently, calculate the number of moles of \(\text{CCl}_4\) present in \(284 \, \text{g}\):

\[\text{Moles of CCl}_4 = \frac{284 \, \text{g}}{154 \, \text{g mol}^{-1}} = \frac{284}{154} \approx 1.844 \, \text{mol}\] Next, compute the heat required for vaporization using the formula:

\[\text{Heat required} = \text{moles} \times \Delta H_v = 1.844 \, \text{mol} \times 30.5 \, \text{kJ mol}^{-1} \approx 56.222 \, \text{kJ}\]

This value, when rounded, is approximately \(56 \, \text{kJ}\).