Identify the clue from reaction with concentrated $H_2SO_4$

The sodium salt of an organic acid $X$ gives effervescence with conc. $H_2SO_4$. This suggests that on acidification, a gaseous product is formed.

Use the clue with acidified aqueous $CaCl_2$

$X$ reacts with acidified aqueous $CaCl_2$ to give a white precipitate.

A well-known test is: $C_2O_4^{2-} + Ca^{2+} \rightarrow CaC_2O_4\downarrow$

Calcium oxalate is a white precipitate.

Use the clue about decolourising acidic $KMnO_4$

The white precipitate formed decolourises acidic $KMnO_4$. Calcium oxalate / oxalic acid is a reducing agent and decolourises acidified permanganate: $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$

This strongly indicates the presence of oxalate ion.

Check option B

Option B is sodium oxalate, $Na_2C_2O_4$.

With conc. $H_2SO_4$: $Na_2C_2O_4 + H_2SO_4 \rightarrow H_2C_2O_4 + Na_2SO_4$

Oxalic acid formed further decomposes on heating with conc. $H_2SO_4$ to give gases such as $CO$ and $CO_2$, causing effervescence.

With acidified $CaCl_2$: $Na_2C_2O_4 + CaCl_2 \rightarrow CaC_2O_4\downarrow + 2NaCl$

The precipitate $CaC_2O_4$ decolourises acidic $KMnO_4$.

Reject other options

• A: $CH_3COONa$
  Gives acetic acid on acidification; no characteristic white precipitate with $CaCl_2$ that decolourises $KMnO_4$.

• C: $C_6H_5COONa$
  Gives benzoic acid on acidification; not consistent with the $CaCl_2$ test or permanganate decolourisation.

• D: $HCOONa$
  Formate can reduce $KMnO_4$, but it does not give the characteristic white precipitate with $CaCl_2$ like oxalate does.

Conclusion

Therefore, $X$ is: $\boxed{Na_2C_2O_4}$

So the correct option is B.