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The solubility for a salt for all values of temperature is difficult to obtain, most values are given for discrete temperatures values. I will take the solubility of lead nitrate at 60 degree C. and 30 degree C instead of 54 degree C and 27 degree C to work out the given problem.
At 60 degree C the solubility of lead nitrate is 91.6 g/100 g of water. 25 g of concentrated solution has 91.6/4 = 22.9 g of lead nitrate. At 30 degree C the solubility decreases to 63.4 g per 100 g of water. So the amount of salt that can be dissolved at this temperature in 25 g of water is 15.85 g.
The amount of lead nitrate that precipitates when the temperature cools from 60 degree C to 30 degree C is 22.9 - 15.85 = 7.05 g.
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