camphor melts at 179.8 degrees, and it has a particularly high freezing point depression constant.(kf=40 degrees/m). when 0.186g of an organic substance of an unknown molar mass is dissolved in 22.01g of liquid camphor, the freezing point of the mixture is found to be 176.7 degrees. what is the molar mass of the solute?

what i have done is:

179.8-176.7/40 = 3.1/40 = 0.0775mol
then,
0.0775*22.01g = 1.706moles
then,
0.186g/1.705mol = 0.109g/mol
then answer is 110g/mol
where have i gone wrong?

Absolutely not. The answer I have is the answer to stay that way.

The equation for molality is m = moles/kg solvent. You MUST insert kg solvent and not grams.
Then moles = grams/molar mass or rearrange to molar mass = grams/moles. There is no "times it by 1000) here at all. The answer is molar mass, not molar mass/1000. I did make a typo at the end. The answer is 1.1 x 10^2, not 1.1 x 10^1 so 110 is correct BUT that is too many s.f. Sorry about the typo.

delta T = Kf*molality

Solve for molality.

molality = moles/kg solvent
Solve for moles.

moles = grams/molar mass
Solve for molar mass.

m = moles/kg solvent. kg solvent is 0.0221, you didn't convert to kg but kept it in grams. I found 108.6 which rounds to 109 BUT to two significant figures it is 1.1 x 10^1

To find the molar mass of the solute, we need to use the freezing point depression formula:

ΔT = Kf * m

Where:
ΔT = Change in freezing point (in degrees)
Kf = Freezing point depression constant (in degrees/m)
m = Molality of the solution (in mol/kg)

First, let's find the change in freezing point (ΔT):
ΔT = Freezing point of the pure solvent - Freezing point of the solution

The freezing point of the pure solvent is given as 179.8 degrees, and the freezing point of the solution is 176.7 degrees:
ΔT = 179.8 - 176.7
ΔT = 3.1 degrees

We know the freezing point depression constant (Kf) is 40 degrees/m. Now, we need to find the molality of the solution (m). Molality is calculated by dividing the moles of solute by the mass of the solvent in kg.

First, let's calculate the moles of solute:
moles = mass / molar mass

The mass of the organic substance is given as 0.186g. Let's assume the molar mass of the solute as M.

moles = 0.186g / M

The mass of the solvent (camphor) is given as 22.01g. We need to convert it to kg:
mass of solvent = 22.01g = 22.01g / 1000g/kg = 0.02201kg

Now, let's calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
m = (0.186g / M) / 0.02201kg
m = (0.186 / M) / 0.02201

Now, we can substitute the values into the freezing point depression equation to find the molar mass (M):

ΔT = Kf * m
3.1 = 40 * [(0.186 / M) / 0.02201]

Simplifying the equation:
3.1 = (0.186 / M) * (40 / 0.02201)
3.1 = (0.186 * 40) / (M * 0.02201)

To isolate the Molar mass (M), we can rearrange the equation:
M = (0.186 * 40) / (3.1 * 0.02201)

Now, calculate the value of M:
M = (0.186 * 40) / (3.1 * 0.02201)
M ≈ 152

Therefore, the molar mass of the solute is approximately 152 g/mol.

so doing it your way you then habe to times it by 1000 to get it back to g/mol.