Acetic Acid Molarity In Vinegar

Glacial acetic acid is a clear colorless liquid. A 99.7% (w/due west) full-bodied glacial acetic acid can exist obtained from different suppliers.
A 99.7% (w/w) glacial acetic acrid means that 100 m of glacial acetic acrid contains 99.vii g of acetic acid.
The density of glacial acetic acid is ane.049 m/ml at 25°C which means that the weight of the ane ml of glacial acetic acrid is 1.049 grams at 25°C.
Molarity refers to the number of moles of the solute present in 1 liter of solution.
In simple words, i mole is equal to the atomic weight of the substance. For case, 1 mole of acetic acid is equivalent to 60.05 g of acetic acid (molecular weight = 60.05).

Calculation procedure:

Known values
Density of glacial acetic acid	one.049 g/ml
Tooth mass of acerb acid (CH3COOH)	60.05 g/mole
Concentration of glacial acetic acid	99.7% (% by mass, wt/wt)

Stride 1: Calculate the book of 100 grams of glacial acetic acid solution.
Formula:

$D e n southward i t y = \frac{w e i g h t}{five o fifty u grand e}$

$Five o fifty u m e = \frac{west e i grand h t}{d due east n southward i t y}$

$Five o 50 u m e o f 100 chiliad r a m s o f g l a c i a fifty a c e t i c a c i d = \frac{100}{1.049} = 95.329 m fifty$

Note: 99.7% (due west/w) Glacial acerb acrid means 100 yard of glacial acetic acid contains 99.7 grand of acerb acid.

The volume of 100 chiliad of glacial acetic acrid is 95.329 ml. That means 99.7 grams of acetic acid is present in 95.329 ml of glacial acetic acid.

Step two: Calculate how many grams of acetic acid is present in 1000 ml of glacial acetic acid.
95.32 ml of glacial acetic acid contains = 99.7 grams of acetic acid

$1 100050 o f 1000 l a c i a l a c e t i c a c i d w i 50 l c o n t a i north = \frac{99 . vii}{95.329 g r a one thousand s o f a c east t i c a c i d}$

m m 50 o f g l a c i a 50 a c east t i c a c i d west i fifty fifty c o n t a i n = \frac{yard x 99.7}{95.329} = 1045.85 g r a m s o f a c e t i c a c i d

one thousand ml of glacial acetic acid will incorporate 1045.85 grams of acetic acrid.

Step iii: Summate the number of moles of acerb acrid present in 1045.85 grams of acerb acid.
sixty.05 grams of acetic acrid = 1 mole

$1 m r a m o f a c east t i c a c i d = \frac{one}{60.05}$ moles

1045.85 g r a k s = \frac{1045.85 10 one}{60.05} = 17.416 m o fifty e s

Therefore, nosotros can say that 1 liter of glacial acerb acid contains 17.416 moles or in other words molarity of glacial acetic acid (99.7% w/w, density 1.049 g/ml) is equal to 17.416.