Question

Pure acetic acid (hc2h3o2) is a liquid and is known as glacial acetic acid. calculate the molarity of a solution prepared by dissolving 50.00 ml of glacial acetic acid at 25 °c in sufficient water to give 500.0 ml of solution. the density of glacial acetic acid at 25 °c is 1.05 g/ml

Answer 1

In order to find the molarity of the solution, we first require the moles of acetic acid added. For this,we need the mass which is:

Mass = volume * density

Mass = 50 * 1.05
Mass = 52.5 grams


Moles = mass / molecular weight

Moles = 52.5 / 60.05 
Moles = 0.874 mol

Next, we know that the molarity of a solution is:

Molarity = moles / liter
Molarity = 0.874 / 0.5

Molarity = 1.75 M

Answer 2

Answer:

The molarity of a acetic acid solution is 1.530 mol/L.

Explanation:

Mass of the acetic acid = m

Volume of the acetic acid = 50.00 ml

Density of the acetic acid =  d = 1.05 g/mL

Mass = Density × Volume

$m=1.05 g/mL\times 50.00 mL=50.5 g$

Volume of the water = 500.0 mL

Volume of the solution  = V + v

Volume of the final solution = 50.00 mL + 500.0 mL = 550.0 mL = 0.550 L

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

The molarity of a acetic acid solution:

$Molarity=\frac{50.5 g}{60g/mol\times 0.550 L}=1.530 mol/L$