1. Tardigrade
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  3. Chemistry
  4. 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:

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Q. 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:

Delhi UMET/DPMTDelhi UMET/DPMT 2004

Solution:

Use the following formula to solve the problem.
$\Delta S=\frac{\Delta H}{T}$
Given, $\Delta S=$ change in entropy
$=28.8 J K^{-1}$ $=28.8 \times 10^{-3} k J K^{-1}$
$\Delta H=$ change in enthalpy
$=30.5 k J$ mol $^{-1}$
$r=$ temperature
$\therefore 28.8 \times 10^{-3}=\frac{30.5}{T}$
or $T=\frac{30.5}{28.8 \times 10^{-3}}$
$=1.059 \times 10^{3}=1059 K$