One mole of `NaCl_((s))` on melting absorbed 30.5 kJ of heat and its entropy is increased by 28.8 `JK^-1`. The melting point of NaCl is _________.

Options

A. 1059K
B. 30.5K
C. 28.8K
D. 28800 K

Answer

A

Explanation

`NaCl(s) hArr NaCl(l)`
Given that : `Delta H = 30.5kJ mol^(-1)`
`Delta S = 28.8 JK^(-1) = 28.8 xx 10^(-3) kJ K^(-1)`
By using `Delta T= (Delta H)/(Delta S) = (30.5)/(28.8 xx 10^(-3)) = 1059 K`