1.

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

Answer»

1059 K
30.5 K
28.8 K
28800 K

Solution :`NaCl(s) rarrNaCl(l)`
Given that : `DeltaH=30.5 kJ mol^(-1)`
`DeltaS=28.8 JK^(-1)=28.8xx10^(-3)kJ K^(-1)`
By USING `DeltaS=(DeltaH)/(T)=(30.5)/(28.8xx10^(-3))=1059 K`.


Discussion

No Comment Found