Question

How do I solve this problem? "vinegar is a ${\mathbf{5}}\%$ solution by volume of acetic acid. a ${\mathbf{100}}{\text{ }}{\mathbf{ml}}$ sample contains what volume of concentrated acid and what volume of water? show work."

Answer 1

$5\;mL$ acetic acid
$95\;mL$ water
Then ${\mathbf{5}}\%$ of the vinegar, by volume, is acetic acid, and we have ${\mathbf{100}}{\text{ }}{\mathbf{ml}}$ of vinegar, we have $5{\text{ }}mL$ of vinegar.
Likewise, we have $100\% - 5\% = 95\% \;$ existence water, so we have $95\;mL$ of water.

Complete step by step answer:
Volume-in-volume is an additional rather simple way of relating a solution. We simply define the percent total volume contributed by the liquid solute. As with the other types of formulas recycled in biology, we accept that the solvent is water except some other solvent is definite.
$5{\text{ }}mL$ of vinegar are liquefied in a vinegar solution. Its v/v concentration.
{\text{% conc}}. = \dfrac{{5{\text{ml of vinegar}}}}{{100{\text{ml of solution}}}} \times 100{\text{% }} = 5{\text{% }}

Additional information –
When we define a concentration as a percentage without requiring the type of formula, we suggest that the solution is to be situated made using the weight-in-volume (w/v) method
Perhaps the easiest way to define a solution is in terms of weight-in-weight (w/w). The weight of the solute qualified to the weight of the final solution is defined as a percentage
Vinegar is an acid with its main ingredient, acetic acid. It remains widely used for food conservation and cooking. The acetic concentration for table vinegar is classically $5\% \;$ whereas higher concentration upto $18$ is used as preserving. The word "vinegar" is resulting from vin aigre, meaning "sour wine".

A table vinegar sample covers $5\% \;$ (mass/mass) of acetic acid.
To what volume $10mL$ of the above sample should be diluted to prepare $0.10M\;$ acetic acid solution. (density $1.05gm{L^{ - 1}}$ , Molar mass of $CH3COOH - 60.0gmo{l^{ - 1}}$ )
Let, mass of solution = $100g$
Mass of vinegar = $5g$
${\text{molarity}} = \dfrac{{{\text{moles}}}}{{{\text{volume}}}} = {\text{mass}} \times \dfrac{{{\text{density}}}}{{{\text{molar mass}} \times {\text{mass of solution}}}} \times 1000$
${\text{molarity}} = \dfrac{{5 \times 1.05}}{{60{\text{}} \times 100}} \times 1000 = 0.875{\text{M}}$
Again,
${M_1}{V_1} = {M_{2}}{V_{2}}$

\; \\\ {0.875 \times 10{\text{ }} = {\text{ }}0.10 \times {V_{2}}} \end{array}$$ Thus, $${V_2} = 87.5ml$$ **Note:** It is important to recall to dilute the glacial acetic acid, since the straight strength causes a severe chemical burn if practical to the epithelium.