Question

A student placed 13.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 55.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?

Answer 1

molar mass of glucose = 180 g/mol

moles = mass / molar mass.

hence moles of glucose added = 13.5 g / (180 g/mol) = X mol

those moles in 100.0 mL of solution.

Hence moles in 55.0 mL = (X mol x 55.0 mL) / 100.0 mL

that 55.0 mL was diluted to 0.500 L (500 mL)
Hence glucose moles in 500 mL = moles in 55.0 mL
= (X mol x 55.0 mL) / 100.0 mL

Final 100.0 mL was taken from the diluted solution.
Hence moles in final 100.0 mL = ((X mol x 55.0 mL) / 100.0 mL) x (100.0 mL) / 500.0 mL
= X x 11 / 100
Glucose mass in final 100.0 mL = X x 11 / 100 mol x 180 g/mol
= 1.485 g

Answer 2

there are 1.49 grams of glucose in 100 mL of the final solution

Step-by-step explanation:

First, calculate the initial concentration of glucose. As the concentration is the rate between amount of solute (glucose) and solution, we can express it as grams of glucose per volume of solution:

`initial concentration=Ci=(gGlucose)/(VolumeSolution)`

`Ci=(13.5g)/(100mL) =0.135g/mL`

As we are diluting the solution, we use the equation:

(Initial concentration)(initial volume)=(final concentration)(final volume) or (Ci)(Vi)=(Cf)(Vf)

We need the final concentration of glucose to find the grams of glucose in 100 mL of this solution, from the above equation we have:

`Cf=((Ci)(Vi))/(Vf)`

`Cf=((0.135g/mL)(55.0mL))/(500mL)=0.0149g/mL`

Note that 0.500 L = 500 mL, so the units cancelled each other.

And for the definition of concentration we have:

`C=(gGlucose)/(mLSolution)`

`gGlucose=(C)(mLSolution)=(0.0149g/mL)(100mL)=1.49g`

So, there are 1.49 grams of glucose in 100 mL of the final solution which concentration is 0.0149 g/mL