Question

I need help with Chemistry

Answer 1

What is the (a) molality, (b) freezing point, and (c) boiling point of a solution containing 2.20 g napthalene (C10H8) in 42.2 g of benzene (C6H6)?

Answer 2

I know that molality is 0.41 but how do I get freezing and boiling point?

Answer 3

@taramgrant0543664

Answer 4

This is the equation for freezing point ΔT = i Kf m where T is the change in temp,
i is the van't Hoff factor, Kf is the molal dreezing point depression constant and m is the molality of the solute

Answer 5

I used this equation but I kept getting wrong answer. Can you show mw how?

Answer 6

What numbers are you using for your values?

Answer 7

Kb= 2.53 C/m
Kr = 5.1 C/m

Answer 8

Your first step is to convert your grams of naphthalene and benzene into moles

Answer 9

So 2.20g *(1 mol/128.17g/mol)=?
42.2 * (1mol/78.11g/mol)=?

Answer 10

Here I'm sorry I have to go but I found a link that goes through the steps for you Im sorry but I hope it's clear.
http://chemed.chem.purdue.edu/genchem/probsolv/colligative/kf1.3.html

Answer 11

I am still confused.

Answer 12

@JFraser

Answer 13

molality is moles of solute per kilogram of solvent, so convert the 2.20g of naphthalene into moles, and the 42.2g of benzene into kilograms

Answer 14

0.017 mol o naphthalene
0.0422 kg of benzene

Answer 15

Good, you already calculate the molality and you said it is 0.41 m
you also need the freezing and boiling temperature of the pure solvent (benzene)

then with the formula ΔT = Kf m you calculate how much the temperature is going to decrease (k=5.12) (freezing point) or increase (k=2.53) (boiling point) and subtract or add that values from the values of freezing and boiling the pure solvent

Answer 16

Can you show me how? I am still confused.

Answer 17

decrease of freezing point
ΔT = Kf m =5.12 C/m x 0.41 m = 2.099 C
Freezing point of pure benzene = 5.5 C
Freezing point of the solution
5.5 C-2.099 C = 3.452 C

for the boiling point
use the same equation ΔT = Kf m
but use the other k=2.53 C/m
Boiling point of pure benzene = 80.1C
Boiling point of the solution
80.1C + (2.53 C/m x 0.41 m)= ???

Answer 18

What is the (a) molality, (b) freezing point, and (c) boiling point of a solution containing 100.0 g of ethylene glycol (C2H6O2) in 150.0 g of water?

M= 10.74

How about this? I tried using this equation and it is wrong answer.

Answer 19

how did you do? what is your pure solvent and what is the molality?

Answer 20

sorry I see the molality is correct 10.74 m

Answer 21

what are your values of k for the water?

Answer 22

I understood the first question. I have another which I am confused.

m= 10.74
Kb for water = 0.512 C/m
Kf for water = 1.86 C/m

Answer 23

now multiply the values of k times the molality

Answer 24

Freezing point 0-(1.86x10.74)= ???
Boiling point 100+ (0.512 x 10.74) = ?????

Answer 25

the freezing point of the solution is below zero is a negative number.
the boiling point is above 100 C