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JEE Main Chemistry Class 11 Solutions Part 1 Questions
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© examsnet.com
Question : 4
Total: 100
1
kg
of
0.75
molal aqueous solution of sucrose can be cooled up to
−
4
°
C
before freezing. The amount of ice (in g) that will be separated out is .......... . (Nearest integer)
[Given,
K
f
(
H
2
O
)
=
1.86
Kkg
mol
−
1
]
[27 Aug 2021 Shift 1]
Your Answer:
Validate
Solution:
Let mass of water (initially present) = x g
Mass of sucrose
=
(
1000
−
x
)
g
Moles of sucrose
=
1000
−
x
342
Molality
=
moles
of
sucrose
mass
of
water
(
initially
)
0.75
=
(
1000
−
x
342
)
(
x
1000
)
x
1000
=
1000
−
x
342
×
0.75
256.5
x
=
10
6
−
1000
x
⇒ x = 795.86 g
Moles of sucrose = 0.5969
New mass of
H
2
O
=
a
kg
Depression in freezing point
Δ
T
f
=
K
f
×
m
4
=
0.5969
a
×
1.86
⇒ a = 0.2775 kg
Ice separated
=
795.86
−
277.5
= 518.3 g ≈ 518 g.
© examsnet.com
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