Question

1. calculate the concentration of glacial acetic acid (containing 99.6% hac) with a density of 1.05 g·ml⁻¹. how many milliliters of glacial acetic acid should be taken to prepare 500 ml of a 0.10 mol·l⁻¹ hac solution?

Explanation:

Step1: Calculate the molarity of glacial acetic acid

The molar mass of $HAc$ ($CH_3COOH$) is $M=(12\times2 + 1\times4+16\times2)\ g/mol = 60\ g/mol$.
Assume we have $1\ L$ of glacial acetic acid. The mass of the solution is $m =
ho V=1.05\ g/mL\times1000\ mL = 1050\ g$.
The mass of $HAc$ in it is $m_{HAc}=1050\ g\times99.6\%$.
The number of moles of $HAc$, $n=\frac{m_{HAc}}{M}=\frac{1050\ g\times99.6\%}{60\ g/mol}\approx17.5\ mol$.
So the molarity of glacial acetic acid, $c_1 = 17.5\ mol/L$.

Step2: Use the dilution - formula

The dilution formula is $c_1V_1 = c_2V_2$.
We know that $c_2 = 0.10\ mol/L$, $V_2=500\ mL = 0.5\ L$, and $c_1 = 17.5\ mol/L$.
We need to find $V_1$, and from $V_1=\frac{c_2V_2}{c_1}$.
Substitute the values: $V_1=\frac{0.10\ mol/L\times0.5\ L}{17.5\ mol/L}\approx0.00286\ L = 2.86\ mL$.

Answer:

$2.86\ mL$