A 1.234 g sample contains only vitamin C (C6H8O6) and sucralose (C12H19Cl3O8)

Question

When the sample is dissolved in water to a total volume of 27.4 mL,27.4 mL, the osmotic pressure of the solution is 5.25 atm5.25 atm at 303 K.303 K. What is the mass percent of vitamin C and sucralose in the sample?

Answer 1

Given data:

1. Total mass of mixture = 1.234 g
2. Volume of solution = 27.4 mL = 0.0274 L
3. Osmotic pressure, Π = 5.25 atm
4. Temperature, T = 303 K
5. Gas constant, R = 0.08206 L·atm·mol⁻¹·K⁻¹
6. Vitamin C (C₆H₈O₆) molar mass = 176.12 g/mol
7. Sucralose (C₁₂H₁₉Cl₃O₈) molar mass = 397.64 g/mol
8. Both are non-electrolytes (i = 1)

Step 1: Calculate total moles of solute using the osmotic pressure equation

Π = MRT

n(total) = ΠV​ / RT

n(total) = (5.25)(0.0274) / (0.08206)(303)

= 0.00579 mol

Step 2: Set up equations for the two components

Let

n1=moles of vitamin C

n2=moles of sucralose

n1​+n2​=0.00579

For each component mass (m) = moles (n) x molar mass

176.12 n1​ + 397.64 n2 ​= 1.234

Step 3: Solve for n1 and n2:

Substitute n2=0.00579−n1:

176.12 n1 ​+ 397.64(0.00579 − n1​) = 1.234

176.12 n1​+ 2.303−397.64 n1​ = 1.234

−221.52 n1​ = − 1.069

n1​ = 0.00483 mol (vitamin C)

n2​ = 0.00579−0.00483 = 0.00096 mol (sucralose)

Step 4: Find masses of each component

m_vitC = 0.00483 (176.12) = 0.850 gm

m_sucralose = 0.00096(397.64) = 0.382 gm

Step 5: Calculate mass percent

%vitamin C = (0.850/1.234) ×100=68.9%

%sucralose = (0.382/1.234) ×100=31.1%

Final Answer:

Vitamin C = 68.9%

Sucralose = 31.1%

Symbol Meaning
Π	Osmotic pressure (in atmospheres, atm)
M	Molarity (mol/L)
R	Gas constant (0.08206 L·atm·mol⁻¹·K⁻¹)
T	Temperature in kelvins (K)
V	Volume of solution in liters (L)
n	Number of moles
m	Mass (in grams, g)