JEE MAIN - Chemistry (2023 - 15th April Morning Shift - No. 20)
$30.4 \mathrm{~kJ}$ of heat is required to melt one mole of sodium chloride and the entropy change at the melting point is $28.4 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ at 1 atm. The melting point of sodium chloride is _______________ K (Nearest Integer)
Answer
1070
Explanation
The heat required to melt one mole of a substance is known as its molar enthalpy of fusion, denoted as $\Delta H_\mathrm{fus}$. For sodium chloride, $\Delta H_\mathrm{fus}$ is $30.4\ \mathrm{kJ/mol}$.
The entropy change at the melting point is given by $\Delta S = \frac{\Delta H_\mathrm{fus}}{T_\mathrm{m}}$, where $T_\mathrm{m}$ is the melting point in Kelvin. Substituting the given values, we get:
$$28.4\ \mathrm{J/K/mol} = \frac{30.4\ \mathrm{kJ/mol}}{T_\mathrm{m}}$$
Solving for $T_\mathrm{m}$, we get:
$$T_\mathrm{m} = \frac{30.4\ \mathrm{kJ/mol}}{28.4\ \mathrm{J/K/mol}} = 1070.4\ \mathrm{K}$$
Rounding off to the nearest integer, the melting point of sodium chloride is $\boxed{1070\ \mathrm{K}}$.
The entropy change at the melting point is given by $\Delta S = \frac{\Delta H_\mathrm{fus}}{T_\mathrm{m}}$, where $T_\mathrm{m}$ is the melting point in Kelvin. Substituting the given values, we get:
$$28.4\ \mathrm{J/K/mol} = \frac{30.4\ \mathrm{kJ/mol}}{T_\mathrm{m}}$$
Solving for $T_\mathrm{m}$, we get:
$$T_\mathrm{m} = \frac{30.4\ \mathrm{kJ/mol}}{28.4\ \mathrm{J/K/mol}} = 1070.4\ \mathrm{K}$$
Rounding off to the nearest integer, the melting point of sodium chloride is $\boxed{1070\ \mathrm{K}}$.
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