Question

A 1.052 g1.052 g sample contains only vitamin C (C6H8O6)(C6H8O6) and sucralose (C12H19Cl3O8).(C12H19Cl3O8). When the sample is dissolved in water to a total volume of 31.1 mL,31.1 mL, the osmotic pressure of the solution is 3.89 atm3.89 atm at 289 K.289 K. What is the mass percent of vitamin C and sucralose in the sample?

Answer 1

Let the mass of Vitamin C be x

then Mass of Sucralose will be (1.052-x)

Molar mass of Vitamin C = 176 g/mol

Molar mass of Sucralose = 397 g/mol

Osmotic Pressure = mRT

3.89 = m * 0.0821 * 289

m = 0.159

molality comes out to be 0.159 m

molality = (number of moles of solute)/(mass of solvent in Kg)

number of moles of solute = number of moles of Vitamin C + number of moles of Sucralose = x/176 + (1.052-x)/397

molality = (x/176 + (1.052-x)/397)/(31.1/1000) = 0.159

Solving the equation we get

x/176 + (1.052-x)/397 = 0.159 * 0.0311

x/176 + (1.052-x)/397 = 0.0049449

x/176 - x/397 = 0.0049449 - 1.052/397 = 0.002295

x[1/176 - 1/397] = 0.002295

x = 0.7256

Mass % of vitamin C in sample = 0.7256/1.052 * 100 = 68.97

Mass % of Sucralose = 100 - 68.97 = 31.03%

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