If a concentrated solution of acetic acid is 99.5 % HC2 H3 O2 and has a density of 1.05 g/mL, what is the concentration of this acid in moles per litre?

[CH₃COOH] = 17.4 M

Explanation:

Acetic acid → CH₃COOH

Molar mass → 60 g/mol

99.5% is percent by mass concentration. It means that 99.5 grams of solute are contained in 100g of solution.

Density → 1.05 g/mL. This data is always referred to solution, not solute!.

We determine solution's volume:

1.05 g/mL = 100 g / V → V = 100 g /1.05 g/mL → 95.2 mL

Now we know, that 99.5 g of acetic acid are contained in 95.2 mL

Let's convert to mmoles → 99.5 g / 60 g/mol = 1.66 moles

We convert solution's volume to L → 95.2 mL . 1L / 1000 mL = 0.0952 L

M (mol/L) = 1.66 mol / 0.0952 L = 17.4 M

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"