Question

Assuming you have 0.075 grams of benzoic acid in 1 mL of water

A) Calculate the moles of benzoic acid.

B) Assuming you used 1 mL of 2 M NaOH calculate the moles of NaOH present in solution.

C) Is the amount of NaOH sufficient to react with all of the benzoic acid?

Answer 1

A)        No of moles of Benzoic acid = Wt. of Benzoic acid/ Mwt of Benzoic acid(122).         

                                  =     0.075/122

                                                    = 1 ×10^-3moles

                

         B. Caliculate the no of moles of 1ml and 2M NaoH.

           No. of moles of NaOH= Molarity × Volume( in liter)

                                       = 2× 10^-3

C) C₆H₅COOH    +   NaOH----------> C₆H₅COONa   + H₂O

1 mole                1 mole                 1 mole(As per equation)

1 ×10^-3            2× 10^-3                     ( How much )                        

1 ×10^-3

Leftover:   0 moles             1 ×10^-3

Still 1 ×10^-3 moles of NaOH REMAINING HENCE NaOH sufficient enough to react with Benzoic acid.