The hydrated salt, $N a_{2} S O_{4} . n H_{2} O$ undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of $n$ will be:

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Solution

The correct option is D 10
44.1 grams $N a_{2} S O_{4}$ at (1 mole $N a_{2} S O_{4}$ /142 g $N a_{2} S O_{4}$ ) = 0.31 mole $N a_{2} S O_{4}$ .
55.9 g $H_{2} O$ at (1 mole $H_{2} O$ / 18.0 g $H_{2} O$ ) = 3.10 mole $H_{2} O$ .
3.10 moles of water molecules are present in .31 moles of $N a_{2} S O_{4}$ , hence 10 moles of water will be present in 1 mole of hydrated salt.

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