Solution, Solute and Solvent Problem 1
Given that the solubility of N H X 4 C l is 3 7 g / 1 0 0 g water at 2 0 ∘ C , calculate the mass of N H X 4 C l in grams in a 1 0 0 0 g N H X 4 C l saturated solution at 2 0 ∘ C .
You may refer Chemistry - Solution, Solute and Solvent .
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2 solutions
1 0 0 0 − x x = 1 0 0 3 7 ⟹ 1 3 7 x = 3 7 0 0 0 ⟹ x ≈ 2 7 0
Assuming that there is no precipitate in the solution,
"The mass of
N
H
4
C
l
should be given according to the solubility".
Now,
Let
x
=
The mass of
water
in the saturated solution (in grams)
Then
1
0
0
0
−
x
=
The mass of
N
H
4
C
l
in the solution (in grams)
Now,
as there is
x
g
of water in the solution,
let us find out the solubility in
x
g
of water.
Solubility
=
3
7
g
/
1
0
0
g
of water
=
0
.
3
7
g
/
1
g
of water
=
0
.
3
7
x
g
/
x
g
of water
Therefore,
by the argument mentioned in the beginning,
1
0
0
0
−
x
=
0
.
3
7
x
⇒ x = 1 . 3 7 1 0 0 0
⇒ x ≈ 7 3 0 g
Finally,
Mass of
N
H
4
C
l
=
1
0
0
0
−
x
=
2
7
0
g
Darn, the problem wording is tricky! I thought it was 3 7 grams of N H 4 C l per 1 0 0 grams of water!
EDIT: sorry I meant 100 grams total.
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It is 3 7 grams of N H 4 C l per 1 0 0 grams of water! What is the problem?
37g per 100g of water, so 37g per 137g of solution :)
Yeah, when I first started learning chemistry, I thought the exact same thing :)
In other words, there are 3 7 g of N H 4 C l in 1 3 7 g of a saturated solution of N H 4 C l . Thus, the mass is 1 3 7 3 7 1 0 0 0 = 2 7 0 g of N H 4 C l ,