Question

g To what volume (in mL) should 5.07 mL of an 6.82 M acetic acid solution be diluted in order to obtain a final solution that is 0.49 M

Answer 1

Answer:

The volume of the solution must be increased to 71.429 millimeters by adding 66.359 millimeters of solute to reduce molarity from 6.82 M to 0.49 M.

Explanation:

The molarity is a unit used for solution that is equivalent to the amount of moles of solute per unit volume of solution. That is:

$M = \frac{n_{st}}{V_{sol}}$

Where:

$n_{st}$ - Amount of moles of solute, measured in moles.

$V_{sol}$ - Volume of the solution, measured in liters.

To reduce the molarity of the solution, more millimeters of solvent should be added. Firstly, the amount of moles of acetic acid inside the 6.82 M solution needs to be determined ($M_{o} = 6.82\,M$ and $V_{sol} = 5.07\times 10^{-3}\,L$):

$n_{st} = M_{o}\cdot V_{sol}$

$n_{st} = (6.82\,M)\cdot (5.07\times 10^{-3}\,L)$

$n_{st} = 0.035\,mol$

Now, the resulting volume of solution after diluting the acetic acid solution is: ($M_{f} = 0.49\,M$ and $n_{st} = 0.035\,mol$):

$V_{sol} = \frac{n_{st}}{M_{f}}$

$V_{sol} = \frac{0.035\,mol}{0.49\,M}$

$V_{sol} = 71.429\times 10^{-3}\,L$

$V_{sol} = 71.429\,mL$ (1 L = 1000 mL)

The amount of solvent needed to dilute the solution is:

$\Delta V_{sol} = 71.429\times 10^{-3}\,L - 5.07\times 10^{-3}\,L$

$\Delta V_{sol} = 66.359 \times 10^{-3}\,L$

$\Delta V_{sol} = 66.359\,mL$ (1 L = 1000 mL)

The volume of the solution must be increased to 71.429 millimeters by adding 66.359 millimeters of solute to reduce molarity from 6.82 M to 0.49 M.