When 1.33 g of a nonpolar solute was dissolved in 50.0 g of phenol, the latter's freezing point was lowered by 1.454°C. Calculate the molar mass of the solute. The kf of phenol is 7.27 (K·kg)/mol.

What is the normal boiling point of an aqueous solution that has a freezing point of -1.04°C?
The kf of water is 1.86 (K·kg)/mol and the kb of water is 0.51 (K·kg)/mol

I am having a diffculty time with coming up with the answer when dealing with freezing and boiling points? Thank you for the help!

To calculate the molar mass of the solute in the first question, you need to use the formula:

∆T = Kf * m * i

Where:
∆T = change in freezing point (in this case, -1.454°C)
Kf = freezing point depression constant of the solvent (7.27°C/m)
m = molality of the solution (mol solute / kg solvent)
i = van't Hoff factor (the number of particles into which the solute dissociates, often assumed to be 1 for non-ionic solutes)

First, we need to calculate the molality (m) of the solution, which is defined as the number of moles of solute per kilogram of solvent:

m = moles of solute / mass of solvent (in kg)

Mass of solvent = 50.0 g = 0.050 kg
Mass of solute = 1.33 g = 0.00133 kg

So, the molality (m) = 0.00133 kg solute / 0.050 kg solvent = 0.0266 mol/kg

Now, let's substitute the given values into the freezing point depression equation:

-1.454°C = (7.27°C/m) * (0.0266 mol/kg) * i

We can now solve for i:

i = -1.454°C / [(7.27°C/m) * (0.0266 mol/kg)] = -1.454°C / (0.194° molality) ≈ -7.49

Since i cannot be negative and generally assumes to be positive for non-ionic solutes, we take the absolute value of i:

i ≈ 7.49

Now, we can use the molar mass formula to calculate the molar mass of the solute:

molar mass (g/mol) = mass (g) / moles

mass = 1.33 g

moles = mass (g) / molar mass (g/mol) = 1.33 g / molar mass (g/mol)

Substituting back into the equation, we get:

-7.49 = (7.27°C/m) * (0.00133 kg) * (1 / molar mass (g/mol))

Now, solve for the molar mass:

molar mass (g/mol) = 7.27°C / (0.00133 kg * 7.49) ≈ 7.27°C / 0.0099667 kg ≈ 729.1 g/mol

Therefore, the molar mass of the solute is approximately 729.1 g/mol.

Now, moving on to the second question about the normal boiling point of an aqueous solution. To calculate the boiling point elevation, we use a similar formula:

∆T = Kb * m * i

Where:
∆T = change in boiling point (unknown)
Kb = boiling point elevation constant of the solvent (0.51°C/m)
m = molality of the solution (unknown)
i = van't Hoff factor (assumed to be 1 for most aqueous solutions)

To find the boiling point elevation (∆T), we rearrange the equation:

∆T = Kb * m * i

Given:
∆T = -1.04°C (because it is a freezing point, it is negative, but boilings points are positive)
Kb = 0.51 (K·kg)/mol (boiling point elevation constant for water)
i = 1 (assuming a non-ionic solute)

Now, substitute the given values into the equation:

-1.04°C = (0.51 (K·kg)/mol) * m * 1

Rearrange the equation to solve for m:

m = -1.04°C / (0.51 (K·kg)/mol) ≈ -2.04 mol/kg

The molality (m) cannot be negative, so we'll take the absolute value:

m ≈ 2.04 mol/kg

We can now substitute the value of m into the boiling point elevation equation to solve for the change in boiling point (∆T):

∆T = (0.51 (K·kg)/mol) * (2.04 mol/kg) * 1 ≈ 2.06°C

To find the normal boiling point, we add this change in boiling point (∆T) to the boiling point of pure water, which is 100.0°C:

Normal boiling point = 100.0°C + 2.06°C ≈ 102.06°C

Therefore, the normal boiling point of the aqueous solution is approximately 102.06°C.

To solve these problems, we will use the formulas related to colligative properties: freezing point depression and boiling point elevation.

First, let's solve the first problem to find the molar mass of the solute:

1. Calculate the molality (m) of the solution using the freezing point depression (ΔTf) and the freezing point constant (Kf) of phenol:

ΔTf = Kf * m

Rearrange the formula to solve for molality:

m = ΔTf / Kf

m = 1.454°C / 7.27 (K·kg)/mol

Convert ΔTf to Kelvin:

ΔTf = 1.454°C + 273.15 = 274.604 K

Now substitute the values and solve for molality:

m = 274.604 K / 7.27 (K·kg)/mol ≈ 37.79 mol/kg

2. Calculate the moles of the solute:

moles of solute = mass of solute / molar mass

molar mass = mass of solute / moles of solute

Given that the mass of the solute is 1.33 g and the molality is 37.79 mol/kg, we can calculate:

molar mass = 1.33 g / (37.79 mol/kg * 0.050 kg)

molar mass ≈ 1.76 g/mol

Therefore, the molar mass of the solute is approximately 1.76 g/mol.

Moving on to the second problem:

To find the boiling point of an aqueous solution, we can use the boiling point elevation formula:

ΔTb = Kb * m

where ΔTb is the boiling point elevation, Kb is the boiling point constant, and m is the molality.

Given that the freezing point of the solution is -1.04°C, we can convert it to Kelvin:

ΔTb = 1.04°C + 273.15 = 274.11 K

Now we can substitute the values into the formula and solve for molality:

274.11 K = 0.51 (K·kg)/mol * m

m = 274.11 K / 0.51 (K·kg)/mol ≈ 537.16 mol/kg

Therefore, the molality of the solution is approximately 537.16 mol/kg.

Next, we need to find the boiling point elevation. Since this value is not given, we cannot determine the normal boiling point of the solution without additional information.

Please provide additional information or specify the concentration or composition of the solution so that we can calculate the normal boiling point.