1) Oblicz zawart. % C,H,O w fruktozie. 2) Oblicz stęż. % fruktozy gdzie na 1 mol fruktozy przypada 10 moli wody.

1) C₆H₁₂O₆ fruktoza Mc = 12 g/mol MH = 1 g/mol MO = 16 g/mol % C = (6 × 12) / (6 × 12 + 12 + 6 × 16) × 100% = 40% % H = 12 / (6 × 12 + 12 + 6 × 16) × 100% = 6,67% % O = 6 × 16 / (6 × 12 + 12 + 6 × 16) × 100% = 53,33% 2) M H₂O = 18 g/mol --> 10 moli H₂O ma mase 180 gram M C₆H₁₂O₆ = 180 g/mol --> 1 mol fruktozy ma mase 180 g Cp = ms / mr-ru × 100% Cp = 180 / (180 + 180) × 100% = 50%

1) C6H12O6--fruktoza mC6H12O6=6*12u+12*1u+6*16u=180u 180g fruktozy-----100% 72g C----------------x% x=40% C 180g fruktozy-----100% 12g H----------------x% x=6,67% H 180g fruktozy-----100% 96g O----------------x% x=53,33% O 2) mC6H12O6=180u m 10H2O =10*18=180u 180g fruktozy-------360g roztworu(180g fruktozy + 180g wody) xg fruktozy----------100g roztworu x=50g fruktozy x=50%