Question

Determine the melting point of an aqueous solution containing 101 mg of saccharin (c7h5o3ns) added to 1.00 ml of water (density of water = 1.00 g/ml, kf = 1.86°c/m).

Answer 1

Answer is:  the melting point of an aqueous solution is -1,023°C.
m(C₇H₅O₃NS) = 101 mg · 0,001 g/mg = 0,101 g.
V(H₂O) = 1,00 mL.
m(H₂O) = d(H₂O) · V(H₂O).
m(H₂O) = 1 g/mL · 1 mL = 1 g = 0,001 kg.
n(C₇H₅O₃NS) = 0,101 g ÷ 183,2 g/mol = 0,00055 mol.
b(C₇H₅O₃NS) = 0,00055 mol ÷ 0,001 kg = 0,55 mol/kg = 0,55 m.
ΔT = Kf · b(C₇H₅O₃NS).
ΔTsolution) = 1,86°C/m · 0,55 m = 1,023°C.

Answer 2

The freezing point of the solution is - 0.74 °c.

First, we have to determine the mass of the water as follows;

Density of water = mass/ volume

Mass = Density of water × volume

Mass = 1.00 g/ml × 1.00 ml = 1.00 g or 0.001 Kg

Number of moles of solute = 101 × 10^-3 g/183 g/mol = 0.0004 moles

Molality = 0.0004 moles/0.001 Kg = 0.4 m

ΔT = K m i

K = freezing constant

m = molality

i = Van't Hoff factor

ΔT = 1.86°c/m × 0.4 m × 1 = 0.74 °c

Freezing point = 0 °c - 0.74 °c = - 0.74 °c

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