Question:

Molar mass of the salt from NaBr, \(NaNO_3\)​, KI and \(CaF_2\)​ which does not evolve coloured vapours on heating with concentrated \(H_2​SO_4\)​ is _____ g \(mol^{−1}\). (Molar mass in g mol−1: Na : 23, N : 14, K : 39, O : 16, Br : 80, I : 127, F : 19, Ca : 40)

Updated On: Nov 4, 2025
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Correct Answer: 78

Approach Solution - 1

The question requires identifying which salt among \(NaBr\), \(NaNO_3\), \(KI\), and \(CaF_2\) does not release colored vapors when heated with concentrated \(H_2SO_4\). Colored vapors typically result from the presence of halogens like bromine or iodine.

Analyzing each: 

  • NaBr: Reacts with \(H_2SO_4\), releasing bromine (\(Br_2\)), which is colored.
  • KI: Reacts with \(H_2SO_4\), releasing iodine (\(I_2\)), which is colored.
  • \(CaF_2\): Reacts with \(H_2SO_4\), producing non-colored vapors as fluoride ions do not lead to colored vapors.
  • NaNO_3: Produces colorless gases (\(NO_2\) is not formed in significant quantities to create colored vapors when heated with \(H_2SO_4\)).

Thus, both \(CaF_2\) and \(NaNO_3\) do not form colored vapors significantly, but only one fits into the provided mass range.

Let's calculate the molar masses:

  • NaNO\(_3\): \(23\ (Na) + 14\ (N) + 3 \times 16\ (O) = 23 + 14 + 48 = 85\ g/mol\)
  • CaF\(_2\): \(40\ (Ca) + 2 \times 19\ (F) = 40 + 38 = 78\ g/mol\)

The computed molar mass of \(CaF_2\) is \(78\ g/mol\), perfectly fitting the range \(78,78\). Therefore, the molar mass of the salt that does not evolve colored vapors is identified clearly.

The solution confirms that the molar mass of \(CaF_2\) is \(78\ g/mol\).

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Approach Solution -2

CaF$_2$ does not evolve any gas with concentrated H$_2$SO$_4$.
\[ \text{NaBr} \rightarrow \text{evolve Br}_2 \]
\[ \text{NaNO}_3 \rightarrow \text{evolve NO}_2 \]
\[ \text{KI} \rightarrow \text{evolve I}_2 \]

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