s block

well, heating them will distinguish

ok thnx...

another quesn....
Colourless salt (A) ->(heating in presence of NaOH) -> (B) (gas )
A -> CaCl2 soln. -> white ppt. (C)
(B) gives white fumes with HCl
(C) decolourises acidified KMnO4
What is (A)? Explain reacns?

Q. A colourless salt (X) has 50% Na2SO3 and 50% H2O . What is the formula of (X) ? How much of SO2 at NTP is obtained when 2.52 g of (X) reacts with excess of dil.H2SO4?

Ans:

Molecular Mass = 126 + (7 x 18) = 252

2nd part :
Na₂SO₃.7H₂O → Na₂SO₃ + 7H₂O

Na₂SO₃ + H₂SO₄ → Na₂SO₄ + H₂O + SO₂

Now applying unitary method,
252gm Na₂SO₃ gives 64gm SO₂ ..
So, 2.52gm Na₂SO₃ gives 0.64gm SO₂..

0.64 gm of SO₂ will be formed..

I think The anser to ur previous question is Ammonium Chloride.

When Ammonium salts r heated in presence of NaOH pungent smelling gas NH3 is released.
When a glass rod dipped in HCl is brought near da mouth of the test the white ammonia fumes intensify.

Wen CaCl2 is added to salt A the acid radical in the salt A is replaced by Cl.

hence the white ppt. is of NH4Cl.

AND chloride salts decolourise acidified KMno4.