Question

8. A 65.0 mL 0.513 mol/l solution of glucose (C6H1206) was mixed with 125.0 mL of

2.33 mol/l glucose solution. What is the molar concentration of the final solution?
Assume the volumes are additive. The molar mass of glucose is 180 g/mol (10
points)

Answer 1

The molar concentration of the final solution is 1.71
`(moles)/(liter)`

Step-by-step explanation:

Molarity is a way of expressing the concentration of solutions and indicates the number of moles of solute dissolved per liter of solution.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution.:

`Molarity (M)=(number of moles of solute)/(volume)`

Molarity is expressed in units (
`(moles)/(liter)`).

Then, the number of moles of solute can be calculated as:

number of moles of solute= molarity* volume

So, in this case, the final concentration can be calculated as:

`Final molarity (M)=(Total number of moles of solute)/(Total volume)`

where, being 65 mL=0.065 L, 125 mL=0.125 L and 190 mL=0.190 L (because 1000 mL= 1 L):

• Total number of moles of solute= 0.065 L*0.513
  `(moles)/(liter)` + 0.125 L*2.33
  `(moles)/(liter)`= 0.033345 moles + 0.29125 moles= 0.324595 moles
• Total volume= 65 mL + 125 mL= 190 mL= 0.190 L

Replacing:

`Final molarity (M)=(0.324595 moles)/(0.190 L)`

Final molarity ≅ 1.71
`(moles)/(liter)`

The molar concentration of the final solution is 1.71
`(moles)/(liter)`