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One mole of NaCl((s)) on melting absorbe...

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 _________.

A

1059K

B

30.5K

C

28.8K

D

28800 K

Text Solution

Verified by Experts

The correct Answer is:
A
`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`
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