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)

Respuesta :

Answer:

The molar concentration of the final solution is 1.71 [tex]\frac{moles}{liter}[/tex]

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.:

[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in units ([tex]\frac{moles}{liter}[/tex]).

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:

[tex]Final molarity (M)=\frac{Total number of moles of solute}{Total volume}[/tex]

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 [tex]\frac{moles}{liter}[/tex] + 0.125 L*2.33 [tex]\frac{moles}{liter}[/tex]= 0.033345 moles + 0.29125 moles= 0.324595 moles
  • Total volume=  65 mL + 125 mL= 190 mL= 0.190 L

Replacing:

[tex]Final molarity (M)=\frac{0.324595 moles}{0.190 L}[/tex]

Final molarity ≅ 1.71 [tex]\frac{moles}{liter}[/tex]

The molar concentration of the final solution is 1.71 [tex]\frac{moles}{liter}[/tex]