A500 mg of oxalic acid (h2c2o4) dissolved in water and titrated with 17.98ml of 0.0600m potassium hydroxide solution. what is the approximate number of mmol of oxalic acid?

The question hasnt clearly mentioned whether to find the number of mmol of oxalic acid before the titration or after the titration so lets find out both

balanced equation for the reaction is

H₂C₂O₄ + 2KOH ---> K₂C₂O₄ + 2H₂O

number of mmol of H₂C₂O₄ initially - 500 mg / 90 g/mol = 5.55 mmol

molar ratio of H₂C₂O₄  to KOH is 1:2

number of KOH moles added - 0.0600 M x 17.98 x 10⁻³ dm³ = 1.08 mmol

since KOH is the limiting reactant and H₂C₂O₄  is in excess,

2 mol of KOH reacts with 1 mol of H₂C₂O₄

therefore 1.08 mmol reacts with - 1/2 x 1.08 = 0.54 mmol of H₂C₂O₄

initially 5.55 mmol of H₂C₂O₄  is present but 0.54 mmol of H₂C₂O₄  reacts with KOH , remaining H₂C₂O₄  moles are 5.55 - 0.54 = 5.01 mmol of H₂C₂O₄

before the titration

number of H₂C₂O₄  moles - 5.55 mmol

after KOH is added

number of H₂C₂O₄  moles - 5.01 mmol

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole