Answer:
The molality is [tex]0.8901m[/tex]
Explanation:
Let's start defining the molality.
[tex]Molality=\frac{MolSolute}{KgOfSolvent}[/tex]
We also know that in terms of masses :
[tex]SoluteMass+SolventMass=SolutionMass[/tex] (I)
Finally, we define the mass percent as :
[tex]MassPercent=\frac{MassOfSolute}{MassOfSolution}.(100)[/tex]
Using the data of the mass percent we find that :
[tex]13.82=\frac{MassOfSolute}{MassOfSolution}.(100)[/tex]
[tex]\frac{MassOfSolute}{MassOfSolution}=0.1382[/tex] ⇒ [tex]MassOfSolution=\frac{MassOfSolute}{0.1382}[/tex] (II)
We know that the molar mass of glucose is [tex]180.16\frac{g}{mol}[/tex]
Therefore, if we use the mass of 1 mole of glucose ([tex]180.16g[/tex]) in (II) ⇒
[tex]MassOfSolution=\frac{180.16g}{0.1382}[/tex]
[tex]MassOfSolution=1303.618g[/tex]
Now, if we use the equation (I) :
[tex]180.16g+SolventMass=1303.618g[/tex]
[tex]SolventMass=1123.458g[/tex]
[tex]1Kg=1000g[/tex] ⇒ [tex]SolventMass=1.1234Kg[/tex]
We find that 1 mole of glucose ([tex]180.16g[/tex] of glucose) are combined with [tex]1.1234Kg[/tex] of solvent to obtain [tex]1303.618g[/tex] of solution which is a 13.82% by mass glucose solution.
If we want to find the molality, we can replaced all the data in the equation of molality :
[tex]Molality=\frac{(1Mol)OfGlucose}{(1.1234Kg)OfSolvent}[/tex]
[tex]Molality=0.8901m[/tex]
We use 1 mol of glucose in the equation (which corresponds to 180.16 g of glucose)
The letter ''m'' is the unit of molality.
