# Chemistry

3.92g of a nondissociating compound are dissolved in 400.g of liquid benzene. The freezing point of the solution is 5.19 degrees celsius. Calculate the molar mass of the compound. (The freezing point of the pure benzene is 5.50 degrees celsius and its molal freezing point depression is 5.12 degrees celsius x kg/mole. (Answer= 1.6 x 10^2 g/mol)

Delta T = Kf mass of solute
--------------------
mole mass solute x kg solvent

1. 👍 0
2. 👎 0
3. 👁 547
1. I find it better to work in steps.
delta T = Kf*molality
Solve for molality

molality = moles solute/kg solvent
Solve for moles solute

moles solute = grams/molar mass
solve for molar mass.

If you still have trouble, post your work on these three steps and someone will look at it for errors.

1. 👍 0
2. 👎 0
2. Delta T = Kf x molality
5.19 C = 5.12 C kg/mol x molality
molality = 1.01 moles/kg

Molality = moles solute/kg solvent
1.01 = moles solute/0.4
moles solute= .404

Moles solute=grams/molar mass
.404=3.92g/molar mass
molar mass= .103

I think I am still doing this wrong.

1. 👍 0
2. 👎 0
3. Your problem is that you are not substituting delta T. Instead, you are substituting the freezing point of the solution.
delta T = (difference between freezing point of the pure solvent and freezing point of the solution). In this case that is 5.50-5.19 = ??
I took a quick look at the other steps. I think those are right if you correct step #1 then go through the rest of the problem.

1. 👍 0
2. 👎 0
4. Oh snap. Thanks so much :D
I knew I was doing something wrong.

1. 👍 0
2. 👎 0
5. how do you figure out the moled without its molecular mass

1. 👍 0
2. 👎 0
6. moles*

1. 👍 0
2. 👎 0
7. you forgot to subtract the final temp with the initial temp. 5.19-5.50= 0.31
then follow the calculations as you did and the molar mass will end up being 163.33

1. 👍 0
2. 👎 0

## Similar Questions

1. ### general chemisty

Calculate the freezing point of a solution of 20.0 g methyl salicylate, C7H6O2, dissolved in 800 g of benzene, C6H6. Kf for benzene is 5.10°C /m and the freezing point is 5.50°C for benzene. A) -1.05°C B) 1.05°C C) 4.45°C D)

2. ### Chemistry

What will be the freezing point of a solution made by dissolving 5.25 g of naphthalene (C10H8) in 100.0 g benzene (C6H6?). The normal freezing point of benzene is 5.5°C and K f (benzene) is 5.12°C / m.

3. ### chemistry

1. Find the molality of the solution prepared by dissolving 0.238g toluene, C7H8, in 15.8g cyclohexane 2. A pure sample of the solvent phenol has a freezing point of 40.85 degrees C. A 0.414 molal solution of isopropyl alcohol was

4. ### CHEM- FP

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

1. ### chemistry

A 1.60 g sample of a mixture of naphthalene, C10H8, and anthracene, C14H10, is dissolved in 20.0 g benzene, C6H6. The freezing point of the solution is 2.81oC. What is the composition as mass percent of the sample mixture?

2. ### Chem

1.10g of an unknown compound reduces the freezing point of 75.22g benzene from 5.53C to 4.92C. What is the molar mass of the compound?

3. ### Chemistry

Thyroxine, an important hormone that controls the rate of metabolism in the body can be isolated from the thyroid gland. If 0.455 g of thyroxine is dissolved in 10.0 g of benzene, the freezing point of the solution could be

4. ### chemistry

(a) Calculate the molar mass of the unknown compound. The freezing point of benzene is 5.48°C, and the Kf of benzene is 5.12 °C/Molal. Show your work

1. ### Chemistry

A) A compound distributes between water (solvent 1) and benzene (solvent 2) with Kp = 2.7. If 1.0g of the compound were dissolved in 100mL of water, how much compound could be extracted by three 10-mL portions of benzene? B) If

2. ### Chemistry 1200

Freezing-point depression can be used to determine the molecular mass of a compound. Suppose that 1.28 g of an unknown molecule were added to 19.9 g of water and the freezing point of the solution determined. If the new freezing

3. ### chemistry

When 0.500 g of an unknown compound was dissolved in 15.0 g benzene, the freezing point depression was determined to be 0.320 C. The molar mass of the unknown compound is ____ (The freezing point depression constant for benzene is

4. ### Chemistry

What is the freezing point of a soluton is which 10.0 g of naphthalene [C10H8(s)] is dissolved in 50.0 g of benzene [C6H6(l)]? Extra info you might need: benzene= 5.12 degrees C m^-1 Answer should be: -2.46