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A sample of lead (II) nitrate dissolved in 173.2 g of water is heated to 50...

A sample of lead (II) nitrate dissolved in 173.2 g of water is heated to 50 C, what is the maximum amount of lead (II) nitrate that can be dissolved?

I know that the answer is about 147 g but do not understand how to get this answer.

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Answer #1

first we have to find the solubility of lead (ii) nitrate at 50 oC from litrature it has been seen that solubility at 50 oC is 0.45 gram lead nitrate / gram of solution

1 gram solution= 0.45 gram lead nitrate+0.55 gram water

solubility with respect to water = 0.45 gram lead nitrate / (1-0.45) gram water

solubility with respect to water= 0.45 gram lead nitrate/0.55 gram water

solubility with respect to water= 0.818 gram lead nitrate/ gram water

so amount of lead nitrate = (0.818 gram lead nitrate/ gram water )*173.2 gram water

amount of lead nitrate= 141.7 gram

please like the answer

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