A solution of sodium sulfate contains 92 g of $N a^{+}$ ions per kilogram of water. The molality of $N a^{+}$ ions in that solution in $m o l k g^{- 1}$ is:

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Solution

The correct option is C 4
Molality $= \frac{no . of moles}{kilogram} = \frac{92}{23} \times 1 = 4$

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A solution of sodium sulfate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in molkg−1 is:

The correct option is C 4Molality = no. of moles kilogram=9223×1=4

A solution of sodium sulfate contains 92 g of $N a^{+}$ ions per kilogram of water. The molality of $N a^{+}$ ions in that solution in $m o l k g^{- 1}$ is:

The correct option is C 4
Molality $= \frac{no . of moles}{kilogram} = \frac{92}{23} \times 1 = 4$