Determine the freezing point of a water solution of fructose (C6H12O6) made by dissolving 92.0 g of fructose in 202 g of water. Answer in units of ◦C.

-4.7°C

Explanation:

In this question, we want to calculate the freezing point of the water solution of fructose.

We proceed as follows;

Firstly, we find the number of moles of the fructose using molar mass

Number of moles = mass/molar mass

mass is 92g , molar mass of fructose is 180g/mol

Number of moles = 92/180 = 0.51 moles

Next thing is to calculate the molality of the solution = number of moles of fructose/mass of water in kg = 0.51/(202/1000) = 0.51/0.202 = 2.525m

Now, we calculate the freezing point depression;

Δ = × m

where refers to the molal freezing point depression of the solvent = 1.86

Δ = 1.86 × 2.525 = 4.7 °C

Since the presence of impurities decrease freezing point and the normal freezing point of water is 0 °C, the freezing point of the solution is 0-4.7 = -4.7 °C

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons