If the density of pure acetic acid is 1.005g/cm, what is it molarity
Density= 1.005g/cm
10cm=1dm
100cm^2=1dm^2
1000cm^3=1dm^3
=({1.005g/1000}dm)^3=(0.001005dm)^3
=1.015075125*10^-9dm^3
Mass = 1.005g
Volume= 1cm
To convert to dm3
1000cm^3=1dm^3
=(1/1000)dm^3
=0.001dm^3
Molar mass of acetic acid(CH3COOH)= ((12*1)+(1*3)+(12*1)+(16*1)+(16*1)+(1*1))
=(12+3+12+16+16+1)g/mol
=60g/mol
Molarity=mole/volume
Mole= mass/molar mass
=1.005g/60g/mol
=0.01675mol
Recall that we are to get molarity
Molarity= 0.1675mol/0.001dm^3
Molarity =16.75moldm^-3
I expect you meant 1.00g g/cc. You will have 1.005 g/cc x 1000 cc = 1005 g acetic acid/L
mols acetic acid 1005/molar mass = 1005/60 = ?
So ?/L = ?M
16.75moles
Well, if the density of pure acetic acid is 1.005g/cm, then I suppose we could say that it's "dense-ity" not to give us the concentration. However, to calculate the molarity (M), we need to know the molar mass of acetic acid. Do you happen to know that, or should I crack a chemistry joke instead?
To calculate the molarity of a solution, we need to know the formula weight and density of the solute. In this case, we're given the density of pure acetic acid (CH3COOH), which is 1.005 g/cm³.
To determine the molarity, we need to convert the density to the mass and then calculate the moles of the solute. Finally, we divide the moles by the volume in liters to get the molarity.
The formula weight of acetic acid is calculated by adding up the atomic masses of each element in the formula: 2(12.01 g/mol) + 4(1.01 g/mol) + 2(16.00 g/mol) = 60.05 g/mol.
To convert the given density (1.005 g/cm³) to mass (g), we need to know the volume of acetic acid. Let's assume it as 1 cm³ for simplicity. So, the mass of acetic acid is 1 cm³ * 1.005 g/cm³ = 1.005 g.
Now, we can calculate the moles of acetic acid by dividing the mass by the formula weight: 1.005 g / 60.05 g/mol ≈ 0.0167 mol.
Finally, to get the molarity, we divide the moles of acetic acid by the volume of the solution in liters. If we assume we have 1 liter of solution, the molarity would be 0.0167 mol / 1 L = 0.0167 M.
Therefore, the molarity of pure acetic acid with a density of 1.005 g/cm³ would be approximately 0.0167 M.