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Stolchlometry III Calculations Based on Concentrations of Solutions
100 g = 5 55 moles
18 mole
What you don't know is x the moles of sucrose needed By definition,
x moles sucrose
mole fraction = 0 0348 = (5 55 moles H 2 O) + (v mole sucrose)
Therefore,
(5 55)(0 0348)
x =
_
= " ^OO mole sucrose
Weight of sucrose needed = (0 200 mole sucrose) (342
g sucrose —J
\
mole sucrose/
= 68 4 g sucrose
Prepare the solution by dissolving 68 4 g sucrose in 100 g water
Commercial Concentrated Solutions
Many solutions can't be made accurately, or at all, by weighing out the solute
and dissolving it in the proper amount of solvent For example, pure substances
such as HC1 and NH 3 are gases, H 2 SO 4 and HNO 3 are fuming hygroscopic
corrosive liquids, NaOH and KOH avidly absorb water and CO 2 from the air
In such circumstances, the customary procedure is to purchase the chemicals
in the form of extremely concentrated solutions, then dilute them to the desired
strength The following problem is typical
PROBLEM.
Commercial concentrated sulfunc acid is labeled as having a density of 1 84 g/ml
and being 96 0% H 2 SO 4 by weight Calculate the molanty of this solution
SOLUTION:
This is a typical conversion problem in which we want to go from grams of
solution per liter to moles of H 2 SO 4 per liter
'l840
g SOlUt '°
n ) (O 960 IH^ /lmoleH 2 S0 4 \ = |g Q moles H 2 SO 4
liter A
g solution/ \ 98 1 g H 2 SO 4 /
liter
- 18 0 M
PROBLEM:
What are (a) the molahty and (b) the mole fraction of the commercial H 2 SO 4
solution in the previous problem
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