Question

A solution of sodium sulfate contains 92 g of na+ ions per kilogram of water. The molarity of na+ ions in that solution in molkg-1 is

Answer 1

Answer : The molality of $Na^+$ ions in the solution is, 4 mole/Kg

Explanation : Given,

Mass of $Na^+$ ion = 92 g

Molar mass of $Na^+$ ion = 23 g/mole

• Molality : It is defined as the number of moles of solute present in one kilogram of solvent.

Formula used :

$Molality=\frac{\text{Moles of solute}}{\text{Mass of solvent in Kg}}$

First we have to calculate the moles of $Na^+$ ion.

$\text{Moles of }Na^+=\frac{\text{Mass of }Na^+}{\text{Molar mass of }Na^+}=\frac{92g}{23g/mole}=4mole$

Now we have to calculate the molality of solution.

$Molality=\frac{4mole}{1Kg}=4mole/Kg$

Therefore, the molality of $Na^+$ ions in the solution is, 4 mole/Kg

Answer 2

Answer:4moles/kg

Explanation:ANSWER ™

I HOPE YOU WILL UNDERSTAND