Question

2. 99.72 g of glucose (C6H12O6) is dissolved in 52.94 mL H2O to produce a final solution volume of 100.00 mL. Calculate the concentration of the glucose in this saturated solution in units of molarity, molality and mass percentage. (Sample Exercise 11.5) (5 pts)

3. What would the freezing point for the saturated glucose solution be, given that Kf for water is 1.86 oC/m? (Sample Exercise 11.7) (3 pts)

4. What would the freezing point be for the aqueous solution if NaCl were substituted for glucose at the same molal concentration? (2 pts)

Answer 1

Question 2.

a)

m = 99.72 g of glucose

V = 52.94 mL of water --> V = 100 mL

Vfinal = 100 mL = 0.1 L

mol of glucose = mass/MW = 99.72/180.1559 = 0.5535 mol of glucose

M = mol/V

M = 0.5535/0.1 = 5.535 M

b)

molality = mol of solute / kg of solvent

assume D of water = 1 G/mL

so

mass of water = 52.94 g

kg of water = 52.94*10-3 kg

molality = mol of solute / kg of solvent = (0.5535 )/(52.94*10^-3) = 10.4552 molal

c)

mass % of glucose = mass of glucose / total mass * 100%

mass % of glucose = 99.72 /(99.72+52.94) * 100%

% mass of glucose = 65.321%