Question

1.)Glucose has a solubility of 125 g /100 mL. What is the approximate concentration of glucose in a saturated solution? How much glucose can be dissolved in 745 mL of water?

Glucose_____M
Glucose_____g

2.) Determine whether the following salts or molecules will have higher or lower solubility in water at higher temperatures.

KCl(s) + H2O(l)----> KCl(aq), ?Hrxn > 0 ____________
NH3(g) + H2O(l)---> NH3(aq), ?Hrxn< 0 ____________

3.) BaF2 has a Ksp of 2.45 x 10-5. What is the solubility of BaF2? What is the F- concentration in a saturated barium fluoride solution?

Solubility of barium fluoride ___________
[F-]= ___________

Answer 1

1) molar mass of glucose = 180 g/mole

thus, moles of glucose in 125 g of it = mass/molar mass of glucose = 125/180 = 0.6945

Now, molar concentration of glucose in saturated solution = moles of glucose per litre of solution = 0.6945/0.1 = 6.945 M

Now, moles of glucose in 745 ml of solution = molar concentration*volume of solution in litres = 6.945*0.745 = 5.174

Thus, mass of glucose in 745 ml of water = moles*molar mass of glucose = 5.174*180 = 931.25 g

2) KCl(s) + H2O(l)----> KCl(aq), ?Hrxn > 0 ;

Since delta Hrkn > 0 ; i.e. the reaction requires heat for dissolution of the salt in water.Therefore, rise in temperature will increase the solubility of KCl in water.
NH3(g) + H2O(l)---> NH3(aq), ?Hrxn< 0    ;

Since delta Hrkn < 0 ; i.e. the reaction dissipates heat for dissolution of the salt in water.Therefore, rise in temperature will decrease the solubility of NH3 in water.

3) BaF₂(s) <--------> Ba²⁺(aq) + 2F^-(aq) ; K_sp = [Ba²⁺]*[F^-]² = 2.45*10^-5

Now, let the molar solubility of BaF₂ = 's' M

Thus, at eqb.[Ba²⁺] = s M & [F^-] = 2s M

Thus., s*(2s)² = 2.45*10^-5

or, 4s³ = 2.45*10^-5

or, s = 0.0183 M

Thus, [F^-] = 2s = 0.0366 M