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JEE Advanced 2020 Full Test 6 Paper 1
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© examsnet.com
Question : 36
Total: 54
E
is an oxidising agent and when a paper dipped in E solution comes in contact with
S
O
2
changes colour from orange to green.
Now, how many are correct about above flow diagram?
(A) Lime is added with
N
a
2
C
O
3
,
keeps the mass porous so that air becomes accessible to all parts reaction mixture and prevents fusion
(B) 'C' is mainly
F
e
2
O
3
(C) Filtrate is a yellow solution.
(D) During acidification a redox reaction is taking place.
(E) D contain 2 salts in solution and when fractional crystallisation occurs salt of lower molecular mass comes out first and is filtered off.
(F) Acidication gives two salts both hydrated. The one which has more water of hydration is less soluble
(G) When
p
H
of solution is increased the colour of solution of E changes from orange to yellow
(H) The oxidising anions present in flow diagram can be distinguished by
A
g
N
O
3
One gives reddish-brown ppt and other gives brick red ppt.
Your Answer:
Validate
Solution:
D
is wrong other statements are correct
This ore is
F
e
C
r
2
O
4
and it is the prep. Of
K
2
C
r
2
O
7
in figure
F
e
C
r
2
O
4
+
O
2
+
N
a
2
C
O
3
→
N
a
2
C
r
O
4
(
yellow
)
+
F
e
2
O
3
+
C
O
2
after acidification we have
N
a
C
r
2
O
7
⋅
2
H
2
O
and
N
a
2
S
O
4
⋅
10
H
2
O
© examsnet.com
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