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aalyn [17]
3 years ago
7

What is the mole fraction, x, of solute and the molality, m (or b), for an aqueous solution that is 19.0% naoh by mass?

Chemistry
1 answer:
Bond [772]3 years ago
5 0
We will assume that the solvent is water. So, if we have 100 grams of the solution, 19 grams will be sodium hydroxide, while the remaining 81 grams will be water.

The molar weight of sodium hydroxide, NaOH, is 40. The molar weight of water is 18. Finding the moles of each:

NaOH:
19 / 40 = 0.475

Water:
81 / 18 = 4.5

Total moles present:
4.5 + 0.475 = 4.975 moles

The mole fraction of NaOH is:
0.475 / 4.975 = 0.0955

The mole fraction of NaOH is 0.0955
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Explanation:
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Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
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So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
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This tells you that the reaction will always produce
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2
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Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
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3
122.55
g
=
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3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L
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