Solubility of a salt X at 20 degree celcius is 40. At 30 degree celcius 15 gms of X forms a saturated solution in 25 gms of water. What would you observe if 50 gms of a solution of X is cooled from 30 to 20 degree celcius
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Answer:
weight of salt = 90 gram,
solubility = 50
sing formula
solubility of solute =
weightofsolventinsaturatedsolution
weightofsoluteinsaturatedsolution
×100
or 50=
weightofsolventinsaturatedsolutio
90gm
×100
so weight of solvent in saturated solution =
50
90gm
×100=180gm
Hence 180 gm of water is required to prepare a saturated solution.
Explanation:
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