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The solubility of barium nitrate at 100°C is 34 g/100 g of H2O and at 0°C is 5.0 g/100g of H2O. If you start with 100 g of Ba(NO3)2 and make a saturated solution in water at 100°C, how much water is required? If the saturated solution is cooled to 0°C, how much Ba(NO3)2 is precipitated out of solution? The precipitated crystals carry along with them on their surface 4 g of H2O per 100 g of crystals.

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    To dissolve 100 g Ba(NO3)2 @ 100 C we need
    100 g H2O x (100 g/34 g) = ?? g H2O.
    When the solution cools to 0 C we get this much Ba(NO3)2 as a ppt.
    5 g Ba(NO3)2 x (?? g from above/100) = xx g Ba(NO3)2. I get something like 15 g for this but you need to do it more accurately than that. Since the crystals carry extra H2O molecules on their surface (4 g/100 g crystals), the 15 g crystals will have an extra
    4 g x (15/100) = 0.6 so the total mass is 15.6 g. Check my thinking.

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