calculate the freezing point of the solution. after mixing these 2 bottles together, set kf of water = 1.86 ° c / m.
bottle 1 contained 0.3 grams of glucose in 1000 grams of water.
the 2nd bottle contains 0.5 mol fructose in 1000 grams of water.

Answer : The freezing point of solution is 273.467 K

Explanation : Given,

Mass of glucose (solute) = 0.3 g

Mass of water (solvent) = 1000 g = 1 kg

Moles of fructose (solute) = 0.5 mol

Mass of water (solvent) = 1000 g = 1 kg

Molar mass of glucose = 180 g/mole

First we have to calculate the moles of glucose.

Now we have to calculate the total moles after mixing.

Now we have to calculate the molality.

Now we have to calculate the freezing point of solution.

As we know that the depression in freezing point is a colligative property that means it depends on the amount of solute.

Formula used :  

where,

= change in freezing point

= temperature of pure water =

= temperature of solution = ?

= freezing point constant of water =

m = molality = 0.251 mole/kg

Now put all the given values in this formula, we get

conversion used :

Therefore, the freezing point of solution is 273.467 K

mg3n2 is known as magnesium nitride

explanation:

i don't think ke is an compound