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

The molarity of a solution prepared is 1.735 M

From the question,

We are to calculate the molarity of a solution prepared by dissolving 50.00 ml of glacial acetic acid in sufficient water to give 500.0 ml of solution.  

First, we will determine the mass of glacial acetic acid present in the 50.00 mL solution

From the question,

The density of glacial acetic acid at 25 °c is 1.05 g/ml

This means, there are 1.05 g of glacial acetic acid in 1 mL of the solution

If, 1mL of the solution contains 1.05 g of glacial acetic acid

Then, 50 mL of the solution will contain 50 × 1.05 g of glacial acetic acid

50 × 1.05 g = 52.5 g

∴ The mass of glacial acetic acid in the 50 mL solution is 52.5 g

Now, we will determine the number of moles of the glacial acetic acid

From the formula

$Number \ of\ moles =\frac{Mass }{Molar\ mass}$

Mass of glacial acetic acid = 52.5 g

Molar mass of glacial acetic acid = 60.052 g/mol

∴ Number of moles of glacial acetic acid = $\frac{52.5}{60.052 }$

Number of moles of glacial acetic acid = 0.86748 moles

Now, for the molarity (that is, concentration) of the solution

From the formula

Number of moles = Concentration × Volume

Then,

$Concentration = \frac{Number \ of \ moles}{Volume}$

Number of moles of glacial acetic acid = 0.86748 moles

Volume of the final solution = 500.0 mL = 0.5 L

∴ Concentration of the solution = $\frac{0.86748}{0.5}$

Concentration of the solution = 1.735 M

Hence, the molarity of a solution prepared is 1.735 M

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