When equimolar quantity of $N a_{2} S O_{4}$ and conc. $H_{2} S O_{4}$ is cooled, the compound formed is :

N a H S O_{4}

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N a_{2} S O_{4}

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N a O_{2}

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None of these

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Solution

The correct option is A $N a H S O_{4}$
When equimolar quantity of $N a_{2} S O_{4}$ and conc. $H_{2} S O_{4}$ is cooled, crystals of $N a H S O_{4}$ are formed.

$N a_{2} S O_{4} + H_{2} S O_{4} \to 2 N a H S O_{4}$
Hence option A is correct.

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H_{2} S O_{4}

413 K

N a_{2} S O_{4}

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