Reducing temperature on dissolved Pb(NO3)2
Assuming that the solution is saturated, meaning that if you add more salt to the solution it will not dissolve, and neglecting eventual evaporation of water, we have a mass of water of about 228g for 171g of salt (assuming solubility of 75g in 100g at 40°C).

Now, when cooling down to 18°C the solubility drops to ~55g / 100g water. This means that in our solution we still have 
$$228~\text{g}~\ce{H2O} \times \frac{55~\text{g}~\ce{Pb(NO3)2}}{100~\text{g}~\ce{H2O}}=125.4~\text{g}~\ce{Pb(NO3)2}\; . $$

The remaining 45.6g of the compound will have precipitated. Whether they form nice crystals or not depends on various other things, see this.

Now, if the solution is not saturated (meaning your salt can still be fully dissolved, but there may be room for more in the solution) the calculation is impossible if you don't know the mass of the water at the beginning.