a solution is prepared using KI which is turned out to be 20% w/w . density of KI is 1.202kg/ml. the molality of the given solution and mole fraction of of solution are respectively ?
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Answer:
(a)Molar mass of KI = 39 + 127 = 166 g mol−1
20%(mass/mass) aqueous solution of KI means 20g of KI is present in 100 g of solution.
That is,
20g of KI is present in (100 − 20) g of water = 80 g of water
Therefore, molality of the solution
=
m
o
l
e
s
o
f
K
I
M
a
s
s
o
f
w
a
t
e
r
i
n
K
g
=
20
166
0.08
=
1.506
m
=
1.51
m
(approximately)
(b) It is given that the density of the solution = 1.202 g mL−1
∴Volume of 100 g solution
=
m
a
s
s
D
e
n
s
i
t
y
=
100
1.202
=
83.19
m
l
=83.19 × 10−3L
Therefore, molarity of the solution
=
20
166
m
o
l
e
83.19
×
10
−
3
L
= 1.45 M
(c)Moles of KI
=
20
166
=
0.12
m
o
l
Moles of water
=
80
18
=
4.44
m
o
l
Therefore, mole fraction of KI
=
m
o
l
e
s
o
f
K
I
moles of KI+ moles of water=0.12
0.12+4.44=0.0263
Explanation:
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