Q4 / 125 · BITSAT 2014 · Chemistry
The enthalpy of sublimation of aluminium is 330 $\mathrm{kJ} / \mathrm{mol}$. Its Ist, II $^{\text {nd }}$ and III $^{\text {rd }}$ ionization enthalpies are 580,1820 and $2740 \mathrm{~kJ}$ respectively. How much heat has too be supplied (in $\mathrm{kJ}$ ) to convert $13.5 \mathrm{~g}$ of aluminium into $\mathrm{Al}^{3+}$ ions and electrons at $298 \mathrm{k}$
- A5470
- Correct answerB2735
- C4105
- D3765
Solution
Heat needed too be supplied
$$
\text { per } m o l=330+580+1820+2740
$$
$=5470 \mathrm{~kJ}$
No. of mols of Al taken $=\frac{13.5}{27}=0.5 \mathrm{~mol}$
Heat required $=0.5 \times 5470 \mathrm{~kJ}=2735 \mathrm{~kJ}$
Hence, the correct answer is option: 'B'.