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JEE Advanced Model Paper 17 with solutions for online practice
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© examsnet.com
Question : 28
Total: 60
A 34.2 % w/w aqueous solution of sucrose has undergone partial inversion according tofollowing first order reaction.
C
12
H
22
O
11
+
H
2
O
→
C
6
H
12
O
6
+
C
6
H
12
O
6
After 10 min boiling point of solution is 108.3°C.
Given boiling point of pure water = 100°C ,
K
b
(
H
2
O
)
- 0.52 Kg
m
o
l
−
1
Which is /are correct statement?
Half life period of reaction is 10 min
Half life period of reaction is 20 min.
Fraction of sucrose inverted is 0.25 after 10 min.
The observed molecular mass is 234 after 10 min.
Validate
Solution:
C
12
H
22
O
11
+
H
2
O
→
C
6
H
12
O
6
+
C
6
H
12
O
6
0.1 → 0 + 0
0.1 - x → x + x
k =
1
t
ln
0.1
0.1
−
x
..... (1)
Δ
T
b
=
k
b
×
1000
w
(
n
S
+
n
G
+
n
F
)
8.3 = 0.52 ×
1000
65.8
(0.1 - x + x + x)
0.1 + x = 1.05 ⇒ x = 0.05
k =
1
10
× ln
0.1
0.1
−
0.05
=
l
m
2
l
n
2
10
×
t
0.5
⇒
t
0.5
- 10 min
M
obs
=
342
×
0.05
+
180
×
0.05
+
180
×
0.05
0.1
−
0.05
+
0.05
+
0.05
= 234
© examsnet.com
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