What is the molarity of #"19.00 cm"^3"# of glacial acetic acid at #"25"^@"C"#?
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.
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
Finally, use the volume of the solution to determine its molarity
Rounded to four sig figs, the answer will be
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
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.
Mass of GA.
Determine the mass of GA using its given volume and its density.
mass = density x volume =
mass of GA =
Moles of GA
Determine the moles of GA by using its mass and molar mass of 60.05 g/mol.
Moles of GA =
Molarity of GA.
Determine the molarity of GA by dividing the mol GA by the liters of the solution.