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

Answer:

The molality is $0.8901m$

Explanation:

Let's start defining the molality.

$Molality=\frac{MolSolute}{KgOfSolvent}$

We also know that in terms of masses :

$SoluteMass+SolventMass=SolutionMass$ (I)

Finally, we define the mass percent as :

$MassPercent=\frac{MassOfSolute}{MassOfSolution}.(100)$

Using the data of the mass percent we find that :

$13.82=\frac{MassOfSolute}{MassOfSolution}.(100)$

$\frac{MassOfSolute}{MassOfSolution}=0.1382$ ⇒ $MassOfSolution=\frac{MassOfSolute}{0.1382}$    (II)

We know that the molar mass of glucose is $180.16\frac{g}{mol}$

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

$MassOfSolution=\frac{180.16g}{0.1382}$

$MassOfSolution=1303.618g$

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 :

$Molality=\frac{(1Mol)OfGlucose}{(1.1234Kg)OfSolvent}$

$Molality=0.8901m$

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.

Answer 2

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

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