Question

What is the molality of a 13.82% by mass glucose solution? the molar mass of c6h12o6 is 180.16 g/mol?

Answer 1

13.82 g / 180.16 g/mol = .07671 moles
.07671 moles / (86.18 g / 1000 g/kg) = .8901 molal

Let me know if you have any further questions!

Answer 2

The molality is
`0.8901m`

Step-by-step explanation:

Let's start defining the molality.

`Molality=(MolSolute)/(KgOfSolvent)`

We also know that in terms of masses :

`SoluteMass+SolventMass=SolutionMass` (I)

Finally, we define the mass percent as :

`MassPercent=(MassOfSolute)/(MassOfSolution).(100)`

Using the data of the mass percent we find that :

`13.82=(MassOfSolute)/(MassOfSolution).(100)`

`(MassOfSolute)/(MassOfSolution)=0.1382` ⇒
`MassOfSolution=(MassOfSolute)/(0.1382)` (II)

We know that the molar mass of glucose is
`180.16(g)/(mol)`

Therefore, if we use the mass of 1 mole of glucose (
`180.16g`) in (II) ⇒

`MassOfSolution=(180.16g)/(0.1382)`

`MassOfSolution=1303.618g`

Now, if we use the equation (I) :

`180.16g+SolventMass=1303.618g`

`SolventMass=1123.458g`

`1Kg=1000g` ⇒
`SolventMass=1.1234Kg`

We find that 1 mole of glucose (
`180.16g` of glucose) are combined with
`1.1234Kg` of solvent to obtain
`1303.618g` 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 :

`Molality=((1Mol)OfGlucose)/((1.1234Kg)OfSolvent)`

`Molality=0.8901m`

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.