What is the freezing point of an aqueous solution of 15.5 g glucose (C6H12O6) dissolved in 150 g of water?

Freezing point of the solution is -1.07°C

Explanation:

The addition of a solute to a solvent decreases its freezing point following the formula:

ΔT = Kf×m×i

Where ΔT is change in temperature (0°C - X); Kf is constant of solvent (1.86°C/m), m is molality (moles glucose per kg solvent) and i is Van't Hoff factor (1 for glucose).

Molality of the solution is:

15.5g glucose -Molar mass 180.156g/mol- contains:

15.5g × (1mol / 180.156g) = 0.086 moles of glucose / 0.150 kg water = 0.574 m

Replacing:

0°C - X = 1.86°C/m × 0.574 × 1

-X = 1.07°C

Freezing point of the solution is -1.07°C

i believe the answer is three water molecules