Vinegar is sold as % acidity which means a weight/volume or w/v% calculation. The density of acetic acid is 1.05 g/ml and water is (effectively) 1.0 g/ml. Recollect (from CHMY121) the %w/v refers to a solution with a certain mass of additive made up to 100 mL with the addition of (in this case) water; or gramsofsolute100mLsolution The molar mass of Acetic Acid is 60.052 g/mol. What is the molarity of CH3COOH in water of a 5% acidity vinegar solution

0.83M is molarity of the solution

Explanation:

In a 5% w/v solution of vinegar there are 5g of acetic acid per 100mL of solution. Or 50g per liter of solution. As molarity is defined as the moles of the solute (Acetic acid) per liter of solution we need to convert the 50g per liter to moles per liter using molar mass of acetic acid:

Molarity:

50g/L * (1mol / 60.052g) = 0.8326moles/L =

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.