Answer:
Mass % Vitamin C: 73.3%
Mass % sucralose: 26.7%
Explanation:
The mass of Vitamin C (a) + Sucralose (b) is 1.167g. Thus, it is possible to write:
a + b = 1.167g (1)
The osmotic pressure formula in solution is:
π = i×M×R×T
Where π is osmotic pressure (3.67atm), i is Van't Hoff factor (1 for Vitamin C and Sucralose),M is molarity of solution (moles / 0.0359L), R is gas constant (0.082atmL/molK) and T is temperature (285K). Replacing:
3.67atm = 1 × moles / 0.0359L × 0.082atmL/molK × 285K
0.00564 = moles of solution (Moles of Vitamin C + Sucralose)
These moles can be written as:
0.00564 mol = a× (1mol / 176.12g) + b×(1mol / 397.64g) (2)
Where 176.12 is molar mass of vitamin C and 397.64 is molar mass of sucralose
Replacing (1) in (2)
0.00564 = (1.167 - b) / 176.12 + b / 397.64
0.000986 = 0.003163 b
0.312g = b. → Mass of sucralose.
Mass of Vitamin C → 1.167g - 0.312g = 0.855g
Thus, Mass percent is:
Mass % Vitamin C: 0.855g / 1.167g ×100 = 73.3%
Mass % sucralose: 0.312g / 1.167g ×100 = 26.7%
Answer:
Explanation:
Mass of sucrose and sucralose = 1.167 g
Molar mass of sucrose = (C6H8O6)
= (12×6) + (8×1) + (6×16)
= 176 g/mol
Molar mass of = C12H19Cl3O8
= (12×12) + (19×1) + (35.5×3) + (16×8)
= 397.5 g/mol
Pressure, P = 3.67 atm
Volume, V = 35.9 ml
Temperature, T = 285 K
Using PV = nRT
nt = (3.67 × 35.9 × 10^-3)/(285 × 0.08205)
= 0.00563 moles.
Molar mass = mass/number of moles
let x be the mass of sucralose
nt = n1 + n2
0.00563 = (1.167 - x)/176 + x/397.5
0.00563 = 0.00663 - 0.0057x + 0.00252x
0.001 = 0.00318x
x = 0.3145 g
Mass % = mass of sample/total mass × 100
Sucralose mass % = 0.3145/1.167 × 100
= 26.95%
Sucrose mass % = (1.167 - 0.3145)/1.167 × 100
= 73.05%
