Assuming complete dissociation, what is the molality of an aqueous solution of KBr whose freezing point is -2.95 ∘C? The molal freezing-point-depression constant of water is 1.86 (∘C⋅kg)/mol.

Molality =

m

Well, well, well. It seems like you need some help with molality, my friend. Alright, let me put on my thinking clown nose and get to work.

Now, molality (m) is defined as the moles of solute divided by the mass of the solvent in kilograms. It is always fun to calculate molality, so let's dive right in.

First, let's find out how much the freezing point is depressed. The formula for freezing point depression is:

∆Tf = Kf * m

Where ∆Tf is the change in freezing point, Kf is the molal freezing-point-depression constant, and m is the molality.

We know that ∆Tf is -2.95 ∘C and Kf is 1.86 (∘C⋅kg)/mol. We just have to rearrange the equation to solve for m, our lovely molality.

So, m = ∆Tf / Kf

Plug in the values:

m = -2.95 ∘C / 1.86 (∘C⋅kg)/mol

And voila, you've got your molality! Just divide those units and you'll have the answer you're looking for. Good luck, my friend!

To find the molality of the KBr solution, we will use the formula:

Molality = (Change in freezing point) / (Freezing-point-depression constant)

Given that the freezing point of the solution is -2.95 ∘C and the freezing-point-depression constant of water is 1.86 (∘C⋅kg)/mol, we can substitute these values into the formula:

Molality = (-2.95 ∘C) / (1.86 (∘C⋅kg)/mol)

Now, we can simplify the units:

Molality = (-2.95 / 1.86) mol/kg

Finally, we can calculate the molality:

Molality ≈ -1.59 mol/kg

Therefore, the molality of the KBr solution is approximately -1.59 mol/kg.

To find the molality of the aqueous solution of KBr, you need to use the formula:

Molality = (m)/(Kf)

Where:
m = molal freezing-point-depression constant
Kf = freezing point depression constant of the solvent (in this case, water)

In the given question, the value of Kf is 1.86 (∘C⋅kg)/mol, and the freezing point of the solution is -2.95 ∘C. Now, all you need to do is substitute these values into the formula to find the molality:

Molality = (m)/(Kf)
Molality = (-2.95 ∘C)/(1.86 (∘C⋅kg)/mol)

To simplify the units, convert the temperature difference from Celsius to Kelvin:

Molality = (-2.95 + 273.15 K)/ (1.86 (∘C⋅kg)/mol)

Perform the calculation:

Molality = -1.79 K / (1.86 (∘C⋅kg)/mol)
Molality ≈ -0.963 mol/kg

Therefore, the molality of the aqueous solution of KBr is approximately -0.963 mol/kg.