A solution of sodium sulphate contains $92$ g of $N a^{+} i o n s$ per kilogram of water. The molality of the $N a^{+}$ ions in the solution in mol $k g^{- 1}$ is:

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Solution

The correct option is C $4$
Here molecular weight of $N a$ is $23 g / m o l$
$n_{N a^{+}} = \frac{92}{23} = 4$

Molality is a number of moles of solute present in per Kg of solvent.

So, molality $= 4$ molal

Hence, the correct option is $C$

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A solution of sodium sulphate contains 92g of Na+ions per kilogram of water. The molality of the Na+ ions in the solution in mol kg−1 is:

The correct option is C 4Here molecular weight of Na is 23g/molnNa+=9223=4Molality is a number of moles of solute present in per Kg of solvent.So, molality =4 molalHence, the correct option is C

A solution of sodium sulphate contains $92$ g of $N a^{+} i o n s$ per kilogram of water. The molality of the $N a^{+}$ ions in the solution in mol $k g^{- 1}$ is:

The correct option is C $4$
Here molecular weight of $N a$ is $23 g / m o l$
$n_{N a^{+}} = \frac{92}{23} = 4$

Molality is a number of moles of solute present in per Kg of solvent.

So, molality $= 4$ molal

Hence, the correct option is $C$