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Byju's Answer

Standard XII

Chemistry

Chemical Reactivity (Group 1)

A compound A ...

Question

A compound (A) is greenish crystalliner salt which gave the following results :
Addition of $B a C l_{2}$ solution to solution of (A) results in the formation of a white ppt. (B), which is insoluble in dil. $H C l .$

(A) :

F e S O_{4} .7 H_{2} O (B) : B a S O_{4}

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(A)

B a S O_{4} (B) : F e S O_{4} .6 H_{2} O

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(A)

B a S O_{4} (B) : F e S O_{4} .7 H_{2} O

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None of these

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Solution

The correct option is A (A) : $F e S O_{4} .7 H_{2} O (B) : B a S O_{4}$
Option (A) is correct.
The compound A is $F e S O_{4} .7 H_{2} O$ . It is greenish crystalline salt.
It reacts with barium chloride to form white ppt of barium sulphate which insoluble in dil $H C l .$
$F e S O_{4} + B a C l_{2} ⟶ B a S O_{4} (w h i t e p p t) + F e C l_{2}$

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Similar questions

Compound (A) is a greenish crystalline salt which gives the following results when tested :

(a) Addition of

B a C l_{2}

solution to a solution of (A) results in the formation of a white precipitate (B) which is insoluble in dilute

H C l

.
(b) On heating, water vapour and two gases (C) and (D) are liberated, leaving a red brown residue (E).
(c) (E) dissolves in

c o n c . H C l

to give a yellow solution (F).
(d) With

H_{2} S

, the solution (F) yields a yellow-white ppt. (G) which when filtrated, leaves a greenish filtrate (H).
(e) The solution (F) gives a green solution with

S n C l_{2}

, and when reacted with

K_{4} [F e (C N)_{6}]

gives a blue ppt.

Identify compound (A) :

Q. A compound (A) is greenish crystalline salt which gave the following results :
On heating (A), water vapours and two oxides of sulfur, (C) and (D) are liberated leaving a red-brown residue (E).
(E) dissolves in warm concentrated HCl to give a yellow solution (F).

With H

_{2}

S, the solution (F) yields a pale yellow ppt. (G), which when filtered, leaves a greenish filtrate (H).

What is compound H?

Q. Hydrated salt (A) +

B a C l_{2} ⟶

white ppt (B) insoluble in

H N O_{3}

(A) (in dil..HCl) +

H_{2} S ⟶

black ppt soluble in 50%

H N O_{3}

(A) +aq

N H_{3}

(excess)

⟶

deep blue solution (C)
(A) +microscomic salt

△ \to

green (D) when hot
(A) +Kl solution

⟶

reddish brown changing to white ppt after hypo is added.
2.495 g of (A) gave 2.33 g of (B). Here (A) is identified as

C u S O_{4} . 5 H_{2} O

.
If true enter 1, else enter 0.

Metal nitrate

(A)

on heating decomposes, leaving a residue

(B)

which goes into the solution with dil.

H C l

. The solution of

(B)

gives white ppt. with ammonium carbonate solution. Ppt.

C

dissolves in dil.

H C l

and solution is treated with potassium chromate to yellow ppt.

(D)

. Solution

V

with dil.

H_{2} S O_{4}

also gives white ppt.

(E)

insoluble in dil.

H C l

and nitric acid. Precipitate

(E)

is part of white pigment lithopone.

What is

(E)

Q. A compound (A) is greenish crystalline salt which gave the following results :
On heating (A), water vapours and two oxides of sulfur, (C) and (D) are liberated leaving a red-brown residue (E).

(E) dissolves in warm concentrated HCl to give a yellow solution (F).

what is F?

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A compound (A) is greenish crystalliner salt which gave the following results : Addition of BaCl2 solution to solution of (A) results in the formation of a white ppt. (B), which is insoluble in dil. HCl.

The correct option is A (A) : FeSO4.7H2O(B):BaSO4Option (A) is correct.The compound A is FeSO4.7H2O. It is greenish crystalline salt.It reacts with barium chloride to form white ppt of barium sulphate which insoluble in dil HCl.FeSO4+BaCl2⟶BaSO4(whiteppt)+FeCl2

A compound (A) is greenish crystalliner salt which gave the following results :
Addition of $B a C l_{2}$ solution to solution of (A) results in the formation of a white ppt. (B), which is insoluble in dil. $H C l .$