Question

What is the molarity of `"19.00 cm"^3"` of glacial acetic acid at `"25"^@"C"`?

Answer 1

C=1.146 M

Explanation:

The molarity of your solution will be 1.146 M.

You need the density of glacial acetic acid, or anhydrous acetic acid, to be able to solve this problem.

Since molarity is defined as moles of solute, in your case glacial acetic acid, divided by liters of solution, you need to know what the mass of dissolved glacial acetic acid is.

The density of glacial acetic acid is listed as `"1.05 g/mL"` (see here: http://www.csudh.edu/oliver/chemdata/acid-str.htm), which means that you dissolve a mass of

`19.00cancel("mL") * "1.05 g"/(1cancel("mL")) = "19.95 g acetic acid"`

SIDE NOTE One cubic centimeter is equal to 1 mL, so I'll just use mL for volume without doing an actual conversion.

Now use acetic acid's molar mass to see how many moles you have

`19.95cancel("g") * "1 mole"/(60.05cancel("g")) = "0.33222 moles"`

Finally, use the volume of the solution to determine its molarity

`C = n/V = "0.33222 moles"/(290.0 * 10^(-3)"L") = "1.14559 M"`

Rounded to four sig figs, the answer will be

`C = color(green)("1.146 M")`

Answer 2

The molarity will be 1.1445 mol/L or 1.1445 M.

Because you were not given the molarity of the glacial acetic acid, you will have to calculate it using its volume, density, and molar mass. This is why your teacher gave you the temperature (density changes with temperature). The density will have to come from a resource. The molecular formula for glacial acetic acid is `"CH"_3"COOH"`, so you can calculate its molar mass, but I looked it up.

The molarity of a substance is the moles of solute per liter of solution. The symbol for molarity is M, and the units are mol/L.
Note: I will abbreviate glacial acetic acid as GA.

GIVEN/KNOWN:
`V_1=19.00"cm^3`
`"density of GA at 25"^("o")"C"=1.049"g/cm"^3`
http://www.sigmaaldrich.com/catalog/product/sial/537020?lang=en&region=US
`"molar mass of GA"="60.05g/mol"`
http://www.sigmaaldrich.com/catalog/product/sial/537020?lang=en&region=US
`V_2=290.0 "cm"^3"=290.0mL=0.2900L`

UNKNOWN:
`M_"GA"`

SOLUTION:

Mass of GA.
Determine the mass of GA using its given volume and its density.
mass = density x volume = `(1.049 g)/(1 cancel(cm^3))xx19.00 cancel(cm^3)` = `"19.931 g"`
mass of GA = `"19.931 g"`

Moles of GA
Determine the moles of GA by using its mass and molar mass of 60.05 g/mol.
`19.931 cancel("g")xx(1 "mol")/(60.05 cancel("g")` = `"0.33191 mol GA"`

Moles of GA = `"0.33191 mol"`

Molarity of GA.
Determine the molarity of GA by dividing the mol GA by the liters of the solution.

`M = "mol/L"`= `"0.33191 mol"/"0.2900 L"= 1.1445 "mol/L"`