in doing this solubility In doing this solubility experiment, you are asked to k

Question

in doing this solubility In doing this solubility experiment, you are asked to keep the temperature of the water below 100. Why is this necessary? Glucose has a solubility of 125 g/100 mL. What is the approximate concentration of Blucose in a saturated solution? How much glucose can be dissolved in 745 mL of water? [glucose] Glucose .Determine whether the following salts or molecules will have higher or lower in water at higher temperatures. KCI(s)+ H20() BaF2 has a K of 2.45 x 10. What is the solubility of BaF2? What is the F concentration in a saturated barium fluoride solution? Solubility of barium fluoride

Explanation / Answer

Solution :-

1)It is important to keep the water temperature below 100 C because the boiling point of water is 100 c therefore at 100 C the liquid water will convert to vapor this will cause the loss of solvent therefore all of the solute may not get dissolved. Or the resulting solution may be supersaturated solution.

2)Given data

Solubility of glucose 125 g per 100 ml water

Molarity = ?

To calculate the molarity we need to calculate the moles of the glucose

Formula to calculate moles is

Moles = mass / molar mass

Moles of glucose = 125 g / 180.156 g per mol = 0.6938 mol

Molarity = moles / volume in liter

                = 0.6938 mol / 0.100 L

                =6.94 M

How much glucose is soluble in 745 ml water

100 ml water = 125 g glucose

745 ml water = ? g glucose

745 ml * 125 g / 100 ml = 931.25 g glucose

Therefore in 745 ml water mass of glucose that can be soluble = 931.25 g glucose

3) Solubility of the KCl(s) will be more at higher temperature

Nh3(g) is in gas form therefore when temperature increases the solubility of the gas decreases.

4) Ksp of BaF2 =2.45E-5

Dissociation equation of the BaF2 is as follows

BaF2   ----------- > Ba^2+ + 2F-

                              X                2x

Ksp equation is as follows

Ksp = [Ba^2+] [F-]^2

Lets put the values in the formula

2.45E-5 = [x][2x]^2

2.45E-5 = 4x^3

2.45E-5 / 4 = x^3

6.125E-6 = x^3

Taking cube root of both sides we get

0.0183 M =x

Therefore the molar solubility of the BaF2 = 0.0183 mol per L

The concentration of the F- in the saturated solution is calculated as follows

[F-] = 0.0183 M * 2 F- / 1 BaF2 = 0.0366 M

Therefore the concentration of the F- in the saturated solution = 0.0366 M

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