Use the following data to explain why NaCl is soluble in water: ∆H = +31 kJmol-1, S(Na+(aq)) = 320.9 JK-1mol-1, S(Cl-(aq)) = 56.5 JK-1mol-1, S(NaCl(s)) = 72.1 JK-1mol-1 Are there any temperatures at which you would not expect NaCl to dissolve?

This is a nice simple calculation question which explores some basic thermodynamic principles and begins to apply them to simple reactions (in this case, solid salt dissolving in water) allowing predictions to be made about the temperature range over which a process is thermodynamically favourable and will therefore occur.
Solution: ∆S = 305.3JK-1mol-1 = 0.3053kJK-1mol-1
Now use ∆G = ∆H - T∆S and the above data to set up a linear equation in T, ∆G = 31.0 - 0.305T. This equation can now be plotted (or just algebraically solved) to see at which ∆G = 0, this the temperature of equilibrium. It turns out to be 101.6K, which is ~ -170 C, the calculation suggests above this temperature the salt is soluble and below it is insoluble in water - of course, it will be a real struggle to get salt to dissolve in water below 273K (even though it is thermodynamically favourable) because the water will be frozen!