Commercially prepared hcl is 38%by mass density is 1.19gm/cc calculate molality of hcl
Answers
38%by mass means 38g hcl is present in 100 gm SOLUTION. so,
molality=moles/mass of SOLVENT(inkg)
mass of solvent is 100-38=62
Answer:
Here's how you can approach such problems.
Explanation:
You know that your solution is 38% w/w hydrochloric acid. This means that every 100 g of solution will contain 38 g of acid.
To make the calculations easier, assume that you're dealing with a 1.00-L sample of stock solution.
Use the solution's density to determine what the mass of this sample would be
1.00
L
⋅
1000
mL
1
L
⋅
1.19 g
1
mL
=
1190 g
Now use the known percent concentration by mass to determine how many grams of hydrochloric acid you'd get
1190
g solution
⋅
38 g HCl
100
g solution
=
452.2 g HCl
To determine the solution's molarity, use hydrochloric acid's molar mass - this will get you the number of moles of acid present in the sample.
452.2
g
⋅
1 mole
36.46
g
=
12.40 moles
Since the sample has a volume of 1.00-L, the molarity will be
C
=
n
V
=
12.40 moles
1.00 L
=
12.4 M
To get the solution's molality, you need to know the mass of water. Since you know the mass of the solution and that of acid, you can write
m
sol
=
m
water
+
m
acid
m
water
=
m
sol
−
m
acid
=
1190
−
452.2
=
737.8 g
The molality will thus be - do not forget to convert the mass of water to kilograms!
b
=
n
m
water
=
12.40 moles
737.8
⋅
10
−
3
kg
=
16.8 molal
Since hydrochloric acid,
H
C
l
, can only release one mole of protons per mole of acid in aqueous solution, the solution's normality will be equal to its molarity.
normality
=
C
f
eq
, where
C
- the solution's molarity;
f
eq
- the equivalence factor, in your case equal to 1.
normality
=
12.40 M
1
=
12.4 N
Explanation: