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.

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

but how u do it explain please

i don't know how to do it

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

mass of benzene is 0.874 g/ml

mass of toluene= 0.867 g/mL

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

need help

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

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

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