An aqueous solution of NaCl freezes at -1.6ºC. At what temperature will it boil? Given that K(b) for water is 0.52 deg m^-1.

delta T = i*Kf*m

dT = 1.6
i = 2 for NaCl
You know Kf.
Solve for m

Then dT = i*Kb*m
Solve for dT and add to 100 C to find new boiling point.

Well, let's boil this down! When we talk about boiling, we're basically asking at what temperature the liquid turns into a gas. In this case, we're dealing with an aqueous solution of NaCl.

When a solute like NaCl is added to water, it raises the boiling point of the solution. This increase is measured by the boiling point elevation constant, which is represented as K(b).

Now, let's calculate the boiling point elevation. The formula for boiling point elevation is:

∆T(b) = K(b) * m

Where ∆T(b) is the boiling point elevation, K(b) is the boiling point elevation constant, and m is the molality of the solution (the number of moles of solute divided by the mass of the solvent in kilograms).

To find the boiling point elevation, we first need to calculate the molality (m). Since we have the freezing point depression, we can figure out the molality using the equation:

∆T(f) = -K(f) * m

Given that the freezing point depression (∆T(f)) is -1.6ºC and the freezing point depression constant (K(f)) for water is 1.86 deg m^-1, we can rearrange the equation to solve for m:

m = ∆T(f) / -K(f)

m = -1.6ºC / -1.86 deg m^-1

m ≈ 0.86 mol/kg

Now, we can plug this value of m into the boiling point elevation equation to find ∆T(b):

∆T(b) = K(b) * m
∆T(b) = 0.52 deg m^-1 * 0.86 mol/kg

∆T(b) ≈ 0.4472ºC

To find the boiling point of the solution, we need to add the boiling point elevation to the normal boiling point of water, which is 100ºC:

Boiling point = Normal boiling point of water + ∆T(b)
Boiling point = 100ºC + 0.4472ºC

Boiling point ≈ 100.4472ºC

So, the aqueous solution of NaCl will boil at approximately 100.4472ºC. Just a tad warmer!

To find the boiling point of a solution, we can use the equation:

ΔTb = Kb * m

Where:
ΔTb is the boiling point elevation (difference between the boiling point of the solution and the boiling point of pure solvent),
Kb is the molal boiling point elevation constant for the solvent (given as 0.52 °C/m for water),
and m is the molality of the solution (moles of solute per kilogram of solvent).

In this case, since we want to find the boiling point (ΔTb), we rearrange the equation:

ΔTb = Kb * m

Rearranging further, we have:

m = ΔTb / Kb

First, calculate the molality of the solution:

ΔTb = Tf - Tb
ΔTb = -1.6ºC - 0ºC = -1.6ºC

m = ΔTb / Kb
m = (-1.6ºC) / (0.52ºC/m)
m = -3.08 m

The molality of the solution is -3.08 m.

Next, we can use the same equation to calculate the boiling point of the solution:

ΔTb = Kb * m

Rearranging the equation for ΔTb, we have:

ΔTb = Tf - Tb
Tf = ΔTb + Tb

Substituting the values we know:

Tf = (-1.6ºC) + (0ºC)
Tf = -1.6ºC

The boiling point of the solution is -1.6ºC.

To find the boiling point of the aqueous solution of NaCl, we can use the equation:

ΔT = K(b) * m

where:
ΔT is the change in boiling point,
K(b) is the molal boiling point constant, and
m is the molality of the solution.

We know the value of K(b) for water, which is 0.52 degrees Celsius per molal (deg C/m). However, we need to convert the temperature from Celsius to Kelvin for calculations, as the Kelvin scale is used in thermodynamics.

First, let's find the change in boiling point (ΔT):

ΔT = -1.6ºC

Next, we need to convert ΔT to Kelvin:

ΔT in Kelvin = ΔT in Celsius + 273.15

ΔT in Kelvin = -1.6 + 273.15 = 271.55 K

Now, we can use the given value of K(b) and the equation to find the molality (m) of the solution:

ΔT = K(b) * m

m = ΔT / K(b)

m = 271.55 K / 0.52 deg C/m

m ≈ 522.60 mol/kg

Finally, we can use the molality (m) to find the boiling point of the solution. We need to use the reverse calculation of the previous step:

ΔT = K(b) * m

Let's plug in the values:

ΔT = 0.52 deg C/m * 522.60 mol/kg

ΔT ≈ 271.55 K

To convert the boiling point back to Celsius, we subtract 273.15:

Boiling point in Celsius = 271.55 K - 273.15 = -1.60ºC

Therefore, the boiling point of the aqueous solution of NaCl is approximately -1.60ºC.