distinguish between nacl and namo3
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one of them is ionic and other is covalent compound
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◼◼ANSWER.. ◼◼
The Salts given are both white crystalline in nature. So, to distinguish which is chloride or nitrate salt of sodium
Take two test tubes and perform the following tests :--
TEST 1:----
▶On adding an Sulphuric acid to these salts as a non volatile acid it will displace a volatile acid with it .
EQUATION :-
In Test tube A:-
⚫
⚛Provided the temperature is less than
200 degrees.
In Test tube B:-
⚫
◽IDENTIFICATION ◽
✳When a glass rod dipped in NH4OH (ammonium hydroxide ) is brought near to test tube A there is no change.
✳While when it's brought near to test tube B dense white fumes of Ammonium chloride (NH4Cl)are formed.
◼Dense white fumes will be formed by Cl-1 radical so the given salt should have been NaCl. Hence, the test tube B have NaCl.
TEST 2:---
▶On adding solution of silver nitrate in both of the test tubes .
EQUATIONS :-
In Test tube A:-
No reaction takes place.
In test tube B:-
◽IDENTIFICATION ◽
There is no change in the test tube A which can only happen when the negative radicals are same . While in test B a white precipitate of AgCl is formed. Which confirms Chloride radical in test tube B.
Therefore, from above two tests we conclused that
⚪Test tube B has chloride radical so, test tube B has sodium chloride
⚪Test tube A has nitrate radical which means it has sodium nitrate.
The Salts given are both white crystalline in nature. So, to distinguish which is chloride or nitrate salt of sodium
Take two test tubes and perform the following tests :--
TEST 1:----
▶On adding an Sulphuric acid to these salts as a non volatile acid it will displace a volatile acid with it .
EQUATION :-
In Test tube A:-
⚫
⚛Provided the temperature is less than
200 degrees.
In Test tube B:-
⚫
◽IDENTIFICATION ◽
✳When a glass rod dipped in NH4OH (ammonium hydroxide ) is brought near to test tube A there is no change.
✳While when it's brought near to test tube B dense white fumes of Ammonium chloride (NH4Cl)are formed.
◼Dense white fumes will be formed by Cl-1 radical so the given salt should have been NaCl. Hence, the test tube B have NaCl.
TEST 2:---
▶On adding solution of silver nitrate in both of the test tubes .
EQUATIONS :-
In Test tube A:-
No reaction takes place.
In test tube B:-
◽IDENTIFICATION ◽
There is no change in the test tube A which can only happen when the negative radicals are same . While in test B a white precipitate of AgCl is formed. Which confirms Chloride radical in test tube B.
Therefore, from above two tests we conclused that
⚪Test tube B has chloride radical so, test tube B has sodium chloride
⚪Test tube A has nitrate radical which means it has sodium nitrate.
Swarup1998:
Great answer! :)
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