MCQ
One mole of $NaCl$ $(s)$ on melting absorbed $30.5\,kJ$ of heat and its entropy is increased by $28.8\,J\,{K^{ - 1}}$. The melting point of $NaCl$ is ............. $\mathrm{K}$
  • $1059$
  • B
    $30.5$
  • C
    $28.8$
  • D
    $28800$

Answer

Correct option: A.
$1059$
(a) $NaCl(s)$ $ \rightleftharpoons $ $NaCl(l)$

Given that : $\Delta H = 30.5\,KJ\,mo{l^{ - 1}}$

$\Delta S = 28.8\,J{K^{ - 1}} = 28.8 \times {10^{ - 3}}\,KJ\,{K^{ - 1}}$

By using $\Delta S = \frac{{\Delta H}}{T}$

$ = \frac{{30.5}}{{28.8 \times {{10}^{ - 3}}}} = 1059\,K$

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