Given solubility of Ba(NO3)2 is
34g/100 H20 at 100oC
5g/100g of H20 AT 0oC
if one starts with 100g of BaNO3)2 and make a saturated solution in water at 100oC, and then cooled to 0oC, how much NO3)2 is crystallised out of solution?
The resultant crystals carry along with them on surface 4g of H20 per 100g of the crystals.
Initial the concentrated solution would be 100g/300g H20 if i do not calculate wrongly.
then how should i proceed?