Freezing point of solution = 49.1 C grams of naphthalene = 2.07 grams of paradic

Question

Freezing point of solution = 49.1 C

grams of naphthalene = 2.07

grams of paradichlorobenzene = 27.00

a.) what is the freezing point depression, Delta T, given that the freezing point for pure paradichlorbenzene is 53.0 C?

I got 3.9 C

b.) using Kf = 7.10 C/m(C kg/mole) for Paradichlorobenzene and m = delta T / Kf, calculate the molarity of the naphthalene.

I got .55 m

c.) using  MM= Kf(number grams solute)/((number kg solvent)(delta T)) calculate the MM of naphthalene

I got ~ 140 g/mol

d.) if part c consistent with the real molar mass of 128 g/ mol?

e.)calculate the percent error that is introduced in the calculated molecular weight, if a 1.0 C error exists in the determination of the delta T? be very specific. show the calculation for 1.0 C higher and 1.0 C lower.

I AM SO SORRY THIS IS SO LONG! PLEASE COMFIRM THAT I HAVE BEEN DOING THIS RIGHT. IF I AM NOT, PLEASE SHOW STEPS. THANK YOU SO MUCH!!!!! :)

Explanation / Answer

1)

Delta T = Tf of pure solvent - Tf of solution : 53 C - 49.1 C = 3.9 C

2)

m = Delta T / k: 3.9 C / 7.1 C/m = 0.55 mole/kg

3.

MM= Kf(number grams solute)/((number kg solvent)(delta T)) = 7.10*2.07/(0.027*3.9) = 139.57 g/mol

4.

Close to the value

Error = |(128 - 139.57)|/128 *100 = 9%

5.

If +1.0 C,

Delta T = 3.9 + 1 = 4.9 C

MM= Kf(number grams solute)/((number kg solvent)(delta T)) = 7.10*2.07/(0.027*4.9) = 111.1g/mol

Error = (128 - 111.1)/128 *100 = 13.2 %

If -1.0 C,

Delta T = 3.9 - 1 = 2.9 C

MM= Kf(number grams solute)/((number kg solvent)(delta T)) = 7.10*2.07/(0.027*2.9) = 187.7g/mol

Error = |(128 - 187.7)|/128 *100 = 46.6 %

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