Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :
[tex]Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}[/tex]
Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:
[tex]0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}[/tex]
[tex]\text{Mass of NaBr}=18.3g[/tex]
Thus, the mass of sodium bromide added should be, 18.3 grams.
