Question

which equation below give the proper units for the molarity of solution of 25.2 grams of acetic acid (molar mass = 60.052 g / mole) in 0.5 l of water? use dimensional analysis of the units to identify the correct equation: m = mole/liter

$o\\ m = \frac{mol_{solute}}{l_{solution}}=\frac{25.2g\\ ch_{3}co_{2}h}{0.500l_{solution}\times\frac{1molch_{3}co_{2}h}{60.052gch_{3}co_{2}h}} = 0.839m$

$o\\ m = \frac{mol_{solute}}{l_{solution}}=\frac{\frac{1molch_{3}co_{2}h}{60.052gch_{3}co_{2}h}}{0.500l_{solution}\times25.2g\\ ch_{3}co_{2}h}} = 0.839m$

$o\\ m = \frac{mol_{solute}}{l_{solution}}=\frac{25.2g\\ ch_{3}co_{2}h\times\frac{1molch_{3}co_{2}h}{60.052gch_{3}co_{2}h}}{0.500l_{solution}} = 0.839m$

$o\\ m = \frac{mol_{solute}}{l_{solution}}=\frac{0.500l_{solution}\times25.2g\\ ch_{3}co_{2}h}{\frac{1molch_{3}co_{2}h}{60.052gch_{3}co_{2}h}} = 0.839m$

Explanation:

Step1: Recall molarity formula

$M=\frac{\text{moles of solute}}{\text{liters of solution}}$

Step2: Calculate moles of acetic - acid

The moles of acetic acid ($n$) is calculated using the formula $n=\frac{m}{M}$, where $m = 25.2$ g and $M = 60.052$ g/mol. So, $n=\frac{25.2\text{ g}}{60.052\text{ g/mol}}$.

Step3: Determine molarity

The volume of the solution $V = 0.5$ L. Molarity $M=\frac{n}{V}=\frac{\frac{25.2\text{ g}}{60.052\text{ g/mol}}}{0.5\text{ L}}=\frac{25.2\text{ g}\times\frac{1\text{ mol}}{60.052\text{ g}}}{0.5\text{ L}}$.

Answer:

$M=\frac{\text{mol solute}}{\text{L solution}}=\frac{25.2\text{ g }CH_3CO_2H\times\frac{1\text{ mol }CH_3CO_2H}{60.052\text{ g }CH_3CO_2H}}{0.500\text{ L solution}} = 0.839\text{ }M$ (the third - option)