A solution contain 38%m of Hcl having density 1.1gm calculate molality and Molarity of the solution and determine the Mole fraction of hcl
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12.61 Molar and 7.96 ml of HCl(38%) needed for 1.0 Liter of 0.10 Molar HCl(aq).
Explanation:
Question 1:
H
C
l
(
38
%
)
=
(
38
g
H
C
l
100
g
Soln
)
=
(
?
moles
HCl
1.0
Liter
Solution
)
=> Moles of HCl in 38 grams =
⎛
⎜
⎜
⎝
38
g
H
C
l
36
(
g
mol
)
⎞
⎟
⎟
⎠
=
1.06
mole
H
C
l
=> Volume of Solution (Liters) containing 1.06 mole of HCl
=
100
g
H
C
l
(
38
%
)
1.19
(
g
ml
)
=
84.03
m
l
H
C
l
(
38
%
)
=
0.084
L
H
C
l
(
38
%
)
=> Molarity HCl(38%) =
1.06
mole
HCl
0.084
L
HCl
Conc
=
12.62
M
o
l
a
r
H
C
l
(
a
q
)
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