Question

4. What is the freezing point of high fructose corn syrup? Assume a content breakdown by mass of 24% water, 68.4% fructose, and 7.6% glucose. Fructose is an isomer of glucose with a molecular formula of C6H1206. (total solution- 100 g water, Freezing-Point Depression Constants(Kf) of water -> textbook, table 11.5) TABLE 11.5 Molal Boiling-Point Elevation Constants (K,) and Freezing-Point Depression Constants (K) for Several Solvents Boiling Freezing Point Point Ke Solvent (°C) (°C . kg/mol) rc-kg/mol) rc) Water (H20) 100.0 0.51 1.86 0 Carbon tetrachloride (CCIA) 76.5 30. 22.99 5.03 4.70 Chloroform (CHCl3) 61.2 3.63 63.5 5.12 2.53 80.1 5.5 Benzene (C&Ho) 3.83 2.34 46.2 Carbon disulfide (CS2) 1.79 116.2 34.5 2.02 Ethyl ether (C4H1o0) 40 179.8 5.95 208.0 Camphor (C10H16O)

Answer 1

Ans 4 :

Fructose and glucose are isomers with the same formula and molar mass

So let us assume the mass of solution be 100 g

Mass of water = (24 x 100 g) / 100 = 24 g = 0.024 kg

So mass of fructose and glucose in it = 100 g - 24 g = 76 g

number of moles = mass / molar mass

= 76 g / 180.156 g/mol = 0.422 mol

Molality = mol of solute / mass of solvent in kg

putting values = 0.422 mol / 0.024 kg

= 17.6 m

Depression in freezing point = kf x m

kf for water = 1.86^oC/m

putting values :

Depression in freezing point = 1.86^oC/m x 17.6 m

=32.7^oC

Freezing point of water = 0^oC

so freezing point of solution = 0^oC - 32.7^oC

= -32.7^oC