Question

What is the weight per volume method to calculate concentration?

Answer 1

Hint :${\displaystyle \frac{Weight}{Volume}}$ percentage concentration is a measurement of the concentration of a solution. ${\displaystyle \frac{Weight}{Volume}}$ percentage concentration is usually called as${\displaystyle \frac{w}{v}}\left(\mathrm{\%}\right)$. Common units for${\displaystyle \frac{w}{v}}\left(\mathrm{\%}\right)$ concentration are${\displaystyle \frac{g}{100ml}}\left(\mathrm{\%}\right)$. The state of being soluble in a given amount of substance is known as solubility which is expressed in terms of concentration also.

Complete Step By Step Answer:
${\displaystyle \frac{Weight}{Volume}}$ is the same as ${\displaystyle \frac{mass}{volume}}$. It is the weight of solute divided by the volume of solution multiplied by $100$: $\mathrm{\%}{\displaystyle \frac{w}{v}}={\displaystyle \frac{\text{weight of solute}}{\text{volume of solution}}}\times 100$
For example: What is the ${\displaystyle \frac{weight}{volume}}$ percentage concentration of $500ml$aqueous acetic acid$\left(C{H}_{3}COOH\right)$ solution containing$10g\text{ C}{\text{H}}_{3}COOH$?
Firstly we write the equation used to calculate the${\displaystyle \frac{weight}{volume}}\left(\mathrm{\%}\right)$:
$={\displaystyle \frac{w}{v}}\mathrm{\%}={\displaystyle \frac{\text{mass of solute}}{\text{volume of solution}}}\times 100$
Now, find the solute: solute$=$acetic acid$\left(C{H}_{3}COOH\right)$
After that, read out the data from the question: mass of solute$C{H}_{3}COOH=10g$
Volume of solution$=500ml$
Put all the values in the formula and solve it:
${\displaystyle \frac{w}{v}}\mathrm{\%}={\displaystyle \frac{10}{500}}\times 100$
$=2\mathrm{\%}$

Note :
${\displaystyle \frac{Weight}{Volume}}$ percent concentrations are characteristic when solids are dissolved in liquids and are mainly used because of the reason that volume is easier to calculate and measure than weight. One more important reason for using this percent concentration is the fact that dilute solutions have a density that’s generally close to$1\text{ gm}{\text{l}}^{-1}$, which makes the volume of the solution expressed in ml numerically equal to the mass of the solution expressed in grams. As a conclusion, in order to calculate a ${\displaystyle \frac{w}{v}}$ percent concentration, we must determine how much quantity of solute is present, it’s mass in grams and how much volume of solution is present in ml. It is mainly used when a dry solute is weighed out and added to a liquid solvent commonly water$\left(H_{2}O\right)$.