Answer:
588.2 mL
Explanation:
- FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)
First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):
- moles = molarity * volume
- 187 mL * 0.692 M = 129.404 mmol Fe⁺²
Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:
- 129.404 mmol Fe⁺² *
= 258.808 mmol KOH
Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:
- volume = moles / molarity
- 258.808 mmol KOH / 0.440 M = 588.2 mL
Answer:- 3333 g of solution.
Some of the question part is missing here. It would be like, "Determine the mass in grams of each NaCl solution that contains 1.5 g of NaCl.
(i) 0.045% NaCl by mass
Solution:- 0.045% NaCl by mass means 0.045 g of NaCl are present in 100 g of solution. 1.5 g of NaCl would be present in how many grams of solution?
We could solve this using proportions...
(0.045/100) = (1.5/X)
0.045(X) = 1.5(100)
0.045X = 150
X = 150/0.045 = 3333
So, 1.5 g of NaCl is present in 3333 g of solution.
Yes it could, but you'd have to set up the process very carefully.
I see two major challenges right away:
1). Displacement of water would not be a wise method, since rock salt
is soluble (dissolves) in water. So as soon as you start lowering it into
your graduated cylinder full of water, its volume would immediately start
to decrease. If you lowered it slowly enough, you might even measure
a volume close to zero, and when you pulled the string back out of the
water, there might be nothing left on the end of it.
So you would have to choose some other fluid besides water ... one in
which rock salt doesn't dissolve. I don't know right now what that could
be. You'd have to shop around and find one.
2). Whatever fluid you did choose, it would also have to be less dense
than rock salt. If it's more dense, then the rock salt just floats in it, and
never goes all the way under. If that happens, then you have a tough
time measuring the total volume of the lump.
So the displacement method could perhaps be used, in principle, but
it would not be easy.