A50 ml sample of vinegar is titrated with 0.774 m naoh(ap). if the triation requires 41.6ml of naoh(ap), what is the concentration of acetic acid in the vinegar

CH3COOH + NaOH > CH3COONa + H2O

The number of mole CH3COOH is equal to the number of mole NaOH in this reaction, so we can write

M(CH3COOH)*V(CH3COOH) = M(NaOH)*V(NaOH)

M(CH3COOH)= M(NaOH)*V(NaOH)/V(CH3COOH)

M(CH3COOH)= 0.774 M*41.6 mL/50 mL ≈ 0.644 M

Answer is M(CH3COOH) ≈ 0.644 M.

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.