Answer: The molality of NaOH solution is 24.69 m
Explanation:
We are given:
Mole fraction of sodium hydroxide = 0.310
This means that 0.310 moles of sodium hydroxide is present in 1.00 moles of solution
Moles of water (solvent) = [1.00 - 0.310] = 0.690 moles
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of water = 0.690 moles
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
[tex]0.690mol=\frac{\text{Moles of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g[/tex]
To calculate the molality of solution, we use the equation:
[tex]Molality=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}[/tex]
Where,
[tex]n_{solute}[/tex] = Moles of solute = 0.310 moles
[tex]W_{solvent}[/tex] = Mass of solvent = 12.42 g
Putting values in above equation, we get:
[tex]\text{Molality of }NaOH=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of }NaOH=24.96m[/tex]
Hence, the molality of NaOH solution is 24.69 m