Question

a solution is prepared by mixing 50.0 mL toluene (C6H5CH3 d=0.867 g/mL) with 125 mL Benzene (C6H6 d=0.874 g/mL). assuming that the volumes add upon mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

Answer 1

http://www.jiskha.com/display.cgi?id=1420516971#1420516971.1420518334

Answer 2

but how u do it explain please

Answer 3

i don't know how to do it

Answer 4

I will get you started with mass percent.
mass percent = (grams solute/mass solution)*100 = %

What mass benzene do you have? Use density to calculate that.

What mass toluene do you have? Use density to calculate that.

mass solution - mass toluene + mass benzene.

Now Substitute into mass percent equation above so that
% toluene = (mass toluene/mass solution()*100 = ?

For the others, lean on the definitions which are as follows:
M = molarity = mols/L solution
m = molality = mols/kg solvent
Xtoluene = mols toluene/total mols
mass = volume x density

Answer 5

mass of benzene is 0.874 g/ml

mass of toluene= 0.867 g/mL

Answer 6

is this right: mass percent: toluene 28.4 %

benezene: 7.6%

mole fraction x C6H5CH3: .251

mole fraction of X C6H6 1.619

molality: 2.3 molal
molarity: 1.43 M is that right? idk

Answer 7

need help

Answer 8

I have bold faced the typo sentence below.

I will get you started with mass percent.
mass percent = (grams solute/mass solution)*100 = %

What mass benzene do you have? Use density to calculate that.

What mass toluene do you have? Use density to calculate that.

<b>mass solution = mass toluene + mass benzene.</b>

Now Substitute into mass percent equation above so that
% toluene = (mass toluene/mass solution()*100 = % toluene

For the others, lean on the definitions which are as follows:
M = molarity = mols/L solution
m = molality = mols/kg solvent
Xtoluene = mols toluene/total mols
mass = volume x density

Answer 9

i figureed is this right DrBOb22

is my ansswers right

mass percent: 43.359/152.6gx100= 28.4%

mass of benzene: 25 mLx .874 g/mL= 109.25 grams

mass of touluene: 50 mLx .867 g/mL= 43.35 grams

molarity: .251 mol/.175 L= 1.43 M

molality: .251 mol/ .10925 kg= 2.30 m
x touluene= .470 mol/1.87 mol= .251 mol

Answer 10

HELP ME WITH THIS ONE WHEN DONE WITH THE OTHER THX
calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25C under a CO2 partial pressure of 0.0003 atm.

c=KP henry's law

k=?

P=0.0003 atm. how to solve

Answer 11

is that right answer

Answer 12

28.4% tolune is right.

Answer 13

what else is right help me

Answer 14

its touluene

Answer 15

what else is right?

Answer 16

The problem doesn't ask for % benzene.
The mole fraction toluene is right (but I obtained 0.252 when rounded).
mole fraction benzene is not but problem doesn't ask for that. However, it is 1-mols fraction toluene = 1-0.252 = ?

M toluene = mols toluene/L soln and soln is 50 mL + 125 mL = 175 mL or 0.175L
Then mol toluee = 43.35/92 = 0.471 so
M = 0.471/0.175 = ?
They don't ask for M benzene but it's done the same way. mols benzene/0.175 = ?
m = mols/kg solvent = 0.471/kg solvent.
mass is approx 153 so m = about 0.471/0.153 = about 3.08

Answer 17

k for CO2 is 3.4E-2 mols/L*atm
c=KP
Substitute p of 3.4E-2 for K and 3E-4 for P, solve for C (in M or mols/L)

Answer 18

a solution is made by dissolving 25 g NaCl in enough water to make 1.0 L solution. Assume that density of the solution is 1.0 g/mL. calculate the mass percent, molarity, molality, and mole fraction of NaCl.

help please

25g/58.44g NaCl=.428 mol

1000 grams of solvent H20

25g/58,44 g x100= 42.8 % mass percent

molarity .428 mol/1.0 L= .428 M

molality= .428 mol/1 Kg= .428 molal is that right?

help please

Answer 19

is that right please help me

Answer 20

25g/58.44g NaCl=.428 mol

1000 grams of solvent H20

25g/58,44 g x100= 42.8 % mass percent
<b> No, you have 25 g NaCl in 1000g which is 2.5 g/100 so it is 2.5%.</b>

molarity .428 mol/1.0 L= .428 M

molality= .428 mol/1 Kg= .428 molal is that right?
<b>The M and m are right.</b>

Answer 21

thanks good night

Answer 22

25g/58.44g NaCl=.428 mol

1000 grams of solvent H20

25g/58,44 g x100= 42.8 % mass percent

molarity .428 mol/1.0 L= .428 M

molality= .428 mol/1 Kg= .428 molal is that right?

help please
ap chemistry - an, Tuesday, January 6, 2015 at 12:47am
is that right please help me
ap chemistry - DrBob222, Tuesday, January 6, 2015 at 12:47am
25g/58.44g NaCl=.428 mol

1000 grams of solvent H20

25g/58,44 g x100= 42.8 % mass percent
No, you have 25 g NaCl in 1000g which is 2.5 g/100 so it is 2.5%.

molarity .428 mol/1.0 L= .428 M

molality= .428 mol/1 Kg= .428 molal is that right?
The M and m are right.

the answer my teacher gave for the molality was .439 m not .428 m

Answer 23

also for the above question the answer for the mole fraction of NaCl is 0.00785 not 0.00766 what i do wrong

moles of NaCl (.428 mol)/ moles of solution (55.9) = 0.00766 what i do wrong?

Answer 24

HELP ME WITH THIS ONE WHEN DONE WITH THE OTHER THX
calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25C under a CO2 partial pressure of 0.0003 atm.

c=KP henry's law

k=?

k for CO2 is 3.4E-2 mols/L*atm
c=KP
Substitute p of 3.4E-2 for K and 3E-4 for P, solve for C (in M or mols/L)
P=0.0003 atm. how to solve

not getting answer of 9.3E-6 M

Answer 25

Yes, and I missed that.
The solution has a mass of 1000 g if the density is 1.0 g/mL but 25g of that is NaCl so the mass solvent is not 1 kg (as I led you to believe) but 1000g-25g = 975g or 0.975 kg.
Then m = mols/kg solvent = 0.428/0.975 = ?
I have answered above that both M and m are right and I'm not going to change all of those duplicate questions and responses. Hope this taks care of it.

I hope this shows you how to get responses. If you show your work we can find the errors faster.

Answer 26

For the (CO2) I used 3.4E-2 for K in mol/L*atm and 29.41 for K in L*atm/mol (just the reciprocal). Your teacher may have used a different K value. Mine came from Wikipedia. A K of 3.25 mol/L*atm will give you 9.75E-6M

Answer 27

tanks