Moles of NH3 in 250 cm ^3 of a 30% solution with specific gravity 0.9 is
Answers
Answer:
Answer:
Here's what I got.
Explanation:
Before doing anything else, make sure that you have a clear understanding of what specific gravity means.
Specific gravity usually refers to the ratio between the density of a given substance and the density of a reference substance at the same temperature. More often than not, that reference substance is water.
SG
=
ρ
substance
ρ
water
Since no mention of temperature was made, you can assume the density of water to be equal to
1.00 g/mL
.
In this context, hydrochloric acid's specific gravity tells you what the density of a specific hydrochloric acid solution is compared with that of water.
More specifically, the density of your hydrochloric acid solution will be equal to
SG
=
ρ
HCl
ρ
water
⇒
ρ
HCl
=
SG
×
ρ
water
ρ
HCl
=
1.08
×
1.00 g/mL
=
1.08 g/mL
So, you know what the density of the solution is. To find its molarity, you need to pick a sample of this solution and figure out how many moles of hydrochloric acid it contains.
Now, you're going to need the percent concentration by mass of hydrochloric acid in this solution, which you can look up here
A hydrochloric acid solution that has a density of
1.08 g/mL
(I assumed room temperature) is about
17
%
w/w
HCl
, that is, you get
17 g
of hydrochloric acid for every
100.0g
of solution.
To make the calculations easier, pick a
1.00-L
sample of this solution. Use its density to find its mass
1.00
L
⋅
1000
mL
1
L
⋅
1.08 g
1
mL
=
1080 g
Next, use its percent concentration by mass to determine how much hydrochloric acid it contains
1080
g solution
⋅
17 g HCl
100
g solution
=
183.6 g
Next, use hydrochloric acid's molar mass to determine how many moles of hydrochloric acid you have in this many grams
183.6
g
⋅
1 mole HCl
36.46
g
=
5.0357 moles HCl
Since molarity is defined as moles of solute, which in your case is hydrochloric acid, divided by liters of solution, and you picked an initial sample of
1.00-L
, it follows that the solution's molarity will be
c
=
n
V
c
=
5.0357 moles
1.00 L
=
5.0357 M
Rounded to three sig figs, the number of sig figs you have for the solution's specific gravity, the answer will be
c
=
5.04 M
Explanation: