Answer:
[tex]\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}[/tex]
Explanation:
Hello.
In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:
[tex]\Delta S=\frac{n*\Delta H}{T}[/tex]
Whereas n accounts for the moles which are computed below:
[tex]n=5.9g*\frac{1mol}{72g} =0.082mol[/tex]
Thus, the entropy turns out:
[tex]\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}[/tex]
Best regards.
