chemistry
posted by Marie on .
Consider 18.0 M H2SO4, if you need to make 250.0 mL of a 3.0 M solution of H2SO4, how would you do this?
18.0 M H2SO4 has a density of 1.84g/mL what is the molality of solution? Mass % of H2SO4 in the solution and the mol fraction of H2SO4 and water in the solution?

#1.
mL1 x M1 = mL2 x M2
Substitute and solve.
250mL x 3.0M = mL2 x 18
#2.
18.0M means 18.0 mols/L
18.0 mols x (98g/mol) = 1764 grams H2SO4/L solution.
The mass of 1L solution is
1.84 g/mL x 1000 mL = 1840 grams H2SO4 + H2O
%H2SO4 = (mass H2SO4/total mass)*100 = (1764/1840)*100 = ?
mols H2SO4 = 1764/98 = ?
mols H2O = (18401764)/18 = ?
total mols = sum
XH2SO4 = mols H2SO4/total mols
XH2O = mols H2O/total mols.
molality = mols/kg solvent. Just substitute the proper numbers.
Answer this Question 
#1 250mL x 3.0 M/18.0M = 41.67mL?
#2 molality of solution???
Mass % 1764/1840x100= 95.87%
Mol fraction...H2SO4. 18/22.2=.81
H2O 4.2/22.2=.18 
#1 looks ok.
#2.
mass % looks ok but if your prof is picky about the number of significant figures I think you need to round that number.
mol fractions look ok except I would have rounded the H2O to 0.19 and not 0.18. In fact, you dropped a 2 from 4.22 and 4.22/22.2 = 0.19
I don't see that you have calculated molality.
That's mols/kg solvent.
You have 18.0 mols and kg H2O = 0.076; therefore, m = 18.0/0.076 = ?