Question

75°C to 35°C the amount of crystals deposited out are much less.
ii) When 800 ml. of a saturated solution of salt 'X' is cooled from 75°C to 35°C a
ing, but when 800 ml. of a saturated solution of salt 'Y' is cooled again from​

Answer 1

Explanation:

Explanation:weight of salt = 90 gram,

Explanation:weight of salt = 90 gram, solubility = 50

Explanation:weight of salt = 90 gram, solubility = 50sing formula

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute =

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolution

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50=

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm ×100

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm ×100so weight of solvent in saturated solution =

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm ×100so weight of solvent in saturated solution = 50

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm ×100so weight of solvent in saturated solution = 5090gm

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm ×100so weight of solvent in saturated solution = 5090gm

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm ×100so weight of solvent in saturated solution = 5090gm ×100=180gm

Explanation:weight of salt = 90 gram, solubility = 50sing formula solubility of solute = weightofsolventinsaturatedsolutionweightofsoluteinsaturatedsolution ×100or 50= weightofsolventinsaturatedsolutio90gm ×100so weight of solvent in saturated solution = 5090gm ×100=180gmHence 180 gm of water is required to prepare a saturated solution.