When a certain amount of MgF2 is added to water, the freezing point lowers by 3.5 oC. What was the molality of the magnesium fluoride? (Kf for water = 1.86 oC/m)

Ah, the fascinating world of chemistry! So, you've got some MgF2 and water, and they're having a "cooling" party. Here's the deal: when you add MgF2 to water, it dissolves and messes with the freezing point.

To figure out the molality, we need to use the formula:

ΔT = Kf * m
ΔT: lowering of the freezing point (in this case, 3.5 oC)
Kf: constant for water (1.86 oC/m)
m: molality (what we're looking for)

Now, let's plug in the numbers we know:

3.5 oC = 1.86 oC/m * m

And now, let's do a little math magic:

m = 3.5 oC / 1.86 oC/m
m ≈ 1.88 mol/kg or 1.88 m

So, the molality of the magnesium fluoride is approximately 1.88 mol/kg. Keep those cool equations coming, and I'll keep the chemistry comedy rolling!

To determine the molality of magnesium fluoride (MgF2), we can use the relationship between the change in freezing point (ΔTf) and the molality of the solute.

The formula for freezing point depression is:

ΔTf = Kf × m × i

Where:
ΔTf = change in the freezing point
Kf = cryoscopic constant for the solvent (water)
m = molality (mol solute/kg solvent)
i = van't Hoff factor (number of particles the solute dissociates into)

Given that the freezing point is lowered by 3.5 oC and the cryoscopic constant (Kf) for water is 1.86 oC/m, we can rearrange the formula and solve for molality (m):

m = (ΔTf) / (Kf × i)

Since magnesium fluoride (MgF2) dissociates into three particles (Mg2+ and two F- ions), the van't Hoff factor (i) is 3.

Plugging in the values, we get:

m = (3.5 oC) / (1.86 oC/m × 3)

Simplifying the equation:

m = 0.628 mol/kg

Therefore, the molality of magnesium fluoride (MgF2) is 0.628 mol/kg.

To find the molality of the magnesium fluoride (MgF2), we need to use the formula:

ΔTf = Kf × m

where:
ΔTf is the freezing point depression (change in freezing point) in degrees Celsius,
Kf is the freezing point depression constant for the solvent (in this case, water), and
m is the molality of the solute (MgF2) in moles of solute per kilogram of solvent.

In this case, the freezing point is lowered by 3.5 oC and the Kf value for water is 1.86 oC/m. We need to calculate the molality (m) of MgF2.

Rearranging the formula, we get:

m = ΔTf / Kf

Substituting the values into the formula:

m = 3.5 oC / 1.86 oC/m

m ≈ 1.88 m

Therefore, the molality of the magnesium fluoride (MgF2) is approximately 1.88 mol/kg.

delta T = i*Kf*m

i = 3, you know the other values, solve for m. By he way, this must be a made up problem because MgF2 isn't that soluble in water. In fact, it is hardly soluble at all.