Question

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

Answer 1

Answer:

The molar concentration of the final solution is 1.71 $\frac{moles}{liter}$

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.:

$Molarity (M)=\frac{number of moles of solute}{volume}$

Molarity is expressed in units ($\frac{moles}{liter}$).

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:

$Final molarity (M)=\frac{Total number of moles of solute}{Total volume}$

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

Replacing:

$Final molarity (M)=\frac{0.324595 moles}{0.190 L}$

Final molarity ≅ 1.71 $\frac{moles}{liter}$

The molar concentration of the final solution is 1.71 $\frac{moles}{liter}$