A solution is prepared by dissolving 20.2 {\rm mL} of methanol ({\rm{CH}}_3 {\rm{OH}}) in 100.0 {\rm mL} of water at 25 ^\circ {\rm C}. The final volume of the solution is 118 {\rm mL}. The densities of methanol and water at this temperature are 0.782 {\rm g/mL} and 1.00 {\rm g/mL}, respectively. For this solution, calculate each of the following.

Question

find 1 ) molality
2) percent by mass
3) moles fraction
4 )mole fraction
5) molarity

Answer 1

20.2 mL CH3OH + 100 mL H2O. Final volume = 118 mL.
1).
Use density to convert 20.2 mL CH3OH to grams. mass = volume*density = 20.2*0.782 = about 15 g(but you need to do it more accurately than that).
Convert 15g to mols. mol = g/molar mass = 15/32 = about 0.5 mol.
Then molality = mols/kg solvent = 0.5/0.1 = ?

2)percent w/w = (grams solute/g soln)*100.
mass solute = about 15g
mass solvent = 100 g
mass solution = 100 + 15 = 115 g
%w/w = (15/115)*100 = ?

3).
mole fraction CH3OH = XCH3OH = mols CH3OH from #1/total mols.
XH2O = mols H2O/total mols.
mols H2O = grams/molar mass = 100/molar mass.

4. Same as 3?
5.M = mols/L soln. You know mols from #1, L soln = 118 mL = 0.118 L.

NOTE: I assume you are same poster under various screen names of aa, anonymous, and others who insist on making this difficult for you to type and for us to answer by continuing to use symbols that, in my opinion, have no place in the question or the subject. 20.2 mL is 20.2 mL no matter how you slice it; writing it as 20.2{\rm
mL} is not only confusing but, in my opinion, is gibberish. You could save yourself a lot of typing time and you could make it lot easier for us to understand and respond by keeping this rubbish out of your posts.