Standard enthalpy of vaporization for CCl4 is 30.5 kJ/mol. Heat required for vaporization of 284 g of CCl4 at constant t

Question

Standard enthalpy of vaporization for CCl4 is 30.5 kJ/mol. Heat required for vaporization of 284 g of CCl4 at constant temperature is:

Accepted Answer

Given: Standard enthalpy of vaporization of $\text{CCl}_4$ is $\Delta H_{\text{vap}} = 30.5 \, \text{kJ mol}^{-1}$ and mass of $\text{CCl}_4$ is $284 \, \text{g}$ .

Find: Heat required for vaporization of $284 \, \text{g}$ of $\text{CCl}_4$ .

First, calculate the molar mass of $\text{CCl}_4$ :

\text{Molar mass} = 1 \times 12 + 4 \times 35.5 = 154 \, \text{g mol}^{-1}

Now calculate the number of moles:

\text{moles of } \text{CCl}_4 = \frac{284}{154} = 1.844 \, \text{mol}

Use the relation:

\text{Heat required} = \text{moles} \times \Delta H_{\text{vap}}

So,

\text{Heat required} = 1.844 \times 30.5 = 56.222 \, \text{kJ}

On rounding, the heat required is $56 \, \text{kJ}$ . Therefore, the correct option is A.

Given: 1 mole of CCl4 requires 30.5 kJ for vaporization. Find: Heat required for 284 g of CCl4. Since molar mass of CCl4 is 154 g/mol, the required heat is directly proportional to mass: Heat = 30.5 284154 = 56.24 kJ Therefore, the required heat is approximately 56 kJ, so the correct option is A.

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