Why crystals appear when a super saturated solution is cooled ?

Answer 1

Are you sure that they do?

#"Supersaturation"# defines a METASTABLE, non-equilibrium condition,... namely that the solvent contains an amount of solute GREATER than that amount which would be equilibrium with undissolved solute. Of course, introduction of a so-called seed crystal, or scratching the sides of the flask, might provide a site of nucleation, and MIGHT cause precipitation of the crystals in solution to give the equilibrium condition, i.e. that the solvent contains an amount of solute that is equal to that amount which would be in equilibrium with UNDISSOLVED solute, in other words a condition of #"saturation"# .

Do you see the distinction I make? I ask because these questions are often very poorly answered at A-level, and even undergrad level, and a little care with your definitions can win easy marks....

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Crystals appear when a supersaturated solution is cooled because cooling reduces the solubility of the solute in the solvent. When a solution becomes supersaturated, it means that it contains more dissolved solute than it can normally hold at a given temperature. This excess solute is typically dissolved due to an increase in temperature or other factors that disrupt the equilibrium between the dissolved and undissolved solute.

As the supersaturated solution is cooled, the solubility of the solute decreases. This decrease in solubility causes the excess solute to precipitate out of the solution in the form of solid crystals. The solute molecules come together and arrange themselves into a crystalline structure as the temperature decreases, following the principles of crystal formation and nucleation.

The appearance of crystals in a supersaturated solution upon cooling is a result of the solute molecules transitioning from a dissolved state to a solid state due to the decrease in temperature. This process allows the solution to return to a state of equilibrium where the concentration of the solute is in balance with its solubility at the lower temperature, resulting in the formation of visible crystals within the solution.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7