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Tardigrade
Question
Chemistry
Arrange the following aqueous solutions in the order of their increasing boiling points (i) 10-4 M NaCl (ii) 10-3 M Urea (iii) 10-3 M MgCl 2 (iv) 10-3 M NaCl
Q. Arrange the following aqueous solutions in the order of their increasing boiling points
(i)
1
0
−
4
M
N
a
Cl
(ii)
1
0
−
3
M
Urea
(iii)
1
0
−
3
M
M
g
C
l
2
(iv)
1
0
−
3
M
N
a
Cl
1562
207
Solutions
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A
(i) < (ii) < (iv) < (iii)
28%
B
(ii) < (i) = (iii) < (iv)
29%
C
(i) < (ii) < (iii) < (iv)
28%
D
(iv) < (iii) < (i) = (ii)
15%
Solution:
(i)
1
0
−
4
M
N
a
Cl
π
=
i
CRT
(
N
a
Cl
⟶
N
a
+
+
C
l
−
)
⇒
2
×
1
0
−
4
RT
(ii)
1
0
−
3
M
Urea
π
=
1
×
1
0
−
3
RT
(iii)
1
0
−
3
MM
g
C
l
2
π
=
3
×
1
0
−
3
RT
M
g
+
2
+
2
C
l
−
(
i
=
3
)
(iv)
1
0
−
3
M
N
a
Cl
π
=
2
×
1
0
−
3
RT
Now i increases
⇒
π
increases
=
Δ
T
b
increases
⇒
(
T
b
−
T
b
∘
)
increases
=
T
b
increases
i.e.
T
b
∝
i
(
i
)
<
(
ii
)
<
(
i
v
)
<
(
iii
)