# ap chemistry

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.

1. 👍
2. 👎
3. 👁
1. http://www.jiskha.com/display.cgi?id=1420516971#1420516971.1420518334

1. 👍
2. 👎
2. but how u do it explain please

1. 👍
2. 👎
3. i don't know how to do it

1. 👍
2. 👎
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

1. 👍
2. 👎
5. mass of benzene is 0.874 g/ml

mass of toluene= 0.867 g/mL

1. 👍
2. 👎
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

1. 👍
2. 👎
7. need help

1. 👍
2. 👎
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.

mass solution = mass toluene + mass benzene.

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

1. 👍
2. 👎
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

1. 👍
2. 👎
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

1. 👍
2. 👎

1. 👍
2. 👎
12. 28.4% tolune is right.

1. 👍
2. 👎
13. what else is right help me

1. 👍
2. 👎
14. its touluene

1. 👍
2. 👎
15. what else is right?

1. 👍
2. 👎
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

1. 👍
2. 👎
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)

1. 👍
2. 👎
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.

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?

1. 👍
2. 👎

1. 👍
2. 👎
20. 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.

1. 👍
2. 👎
21. thanks good night

1. 👍
2. 👎
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?

ap chemistry - an, Tuesday, January 6, 2015 at 12:47am
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

1. 👍
2. 👎
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?

1. 👍
2. 👎
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

1. 👍
2. 👎
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.

1. 👍
2. 👎
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

1. 👍
2. 👎
27. tanks

1. 👍
2. 👎

## Similar Questions

1. ### physics

A mass m = 0.068 kg of benzene vapor (Lv = 3.94x105 J/kg) at its boiling point of 80.1°C is to be condensed by mixing with water at 50.0°C. What is the minimum mass of water required to condense all of the benzene vapor? Assume

2. ### chem

The concentration of a benzene solution prepared by mixing 12.0 g C6H6 with 38.0 g CCl4 is__________ molal.

3. ### Biochem

A buffer solution is prepared by mixing 200 mL of 0.2 M salt solution and 400 mL of a 0.2 M acid solution. What is the concentration of the resulting buffer? ~what is the pKa?

4. ### biochemistry

How do you calculate the PH of a buffer solution prepared by mixing 75 mL of 1.0 M lactic acid and 25 mL of 1.0 M sodium lactate?

1. ### Chemistry

A solution is prepared by mixing 0.12 L of 0.12 M sodium chloride with 0.22L of a 0.19M MgCl2 solution.What volume of a 0.22 M silver nitrate solution is required to precipitate all the Cl- ion in the solution as AgCl ?

2. ### Chemistry

What is the molar mass of toluene if 0.85 g of toluene depresses the freezing point of 100 g of benzene by 0.47 degrees celsius? Kf= 5.12 C/m

3. ### Chemistry

Vapour pressure of benzene and toluene are 0.9bar and 0.85bar. 7.8g of benzene is added to 180g of toluene. To find the vapour pressure of the resultant solution

4. ### Chemistry

A buffer was prepared by mixing 1.00 mol of ammonia 𝑁𝐻3 𝐾𝑎 = 5.8 × 10−10 and 1.00 mol of ammonium chloride 𝑁𝐻4𝐶𝑙 to form an aqueous solution with a total volume of 1.00 liter. To 500 mL of this solution

1. ### 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

2. ### chemistry

At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of pure toluene = 290 torr). The mole fraction of benzene in the vapor above the solution is 0.590.

3. ### Chemistry

An solution of antifreeze is prepared by mixing 23.0mL of ethylene glycol (d = 1.11 g/mL;molar mass = 62.07 g/mol) with 50.0 mL H2O (d = 1.00 g/mL) at 25°C. If the density of the antifreeze solution is 1.07 g/mL, what is its

4. ### Chemistry

calculate the pH ofa solution prepared by mixing 50.0 ml of 0.200 m ch2h3o2 and 50 ml of 0.100 m Naoh.[ka(ch3cooh)=1.8*10^-5]