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____ [38]
3 years ago
6

Additional solute is added to a saturated solution of a solid and a liquid after being heated above room temperature. Once no mo

re of the solute can dissolve, the solution is cooled, and a seed crystal is added. At which of the following points is the solution unsaturated? a. Before the solution is heated, when the concentration of the solute is equal to its solubility in the liquid b. After the solution is heated, but before additional solute is added c. After the solution is heated and more solute is added, but before it is cooled d. Before the solution is heated, when the concentration of the solute is less than its solubility in the liquid
Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

After the solution is heated, but before additional solute is added

Explanation:

An unsaturated solution is a solution that contains less solute than it can normally hold at a given temperature. Hence an unsaturated solution can still dissolve more solute.

When the solution is heated, the saturated cold solution becomes an unsaturated hot solution which is capable of dissolving more solute at this point.

Once more solute is dissolved, the solution becomes saturated again just before it begins to cool since no more solute dissolves in the solution at some point before cooling and addition of seed crystals.

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What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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