Answer:
58.6g/mol
Explanation:
according to dilution principle
no of mole (mol) = conc(M)×vol(L)
for a titration reation involving a monoprotic acid
no of mole of acid =no of mole of base
thus: conc of acid × vol of acid= conc of base ×volume of base
no of mole of base=conc of base × vole of base(NaOH)
=1.0M×0.01348L
=0.01348mol
from the relationship above,
no of mole of acid =no of mole of base
thus no of mole of acid=0.01348mol
However, no of mole =mass\molar mass
molar mass=mass/no of mole
=0.79/0.01348
=58.6g/mol.
Answer:
58.61 g/mol
Explanation:
The acid us monopromatic and hence, 1 mole of it is required to react with  mole of NaOH in order to reach the equivalence point.
In order to detrmine the concentration of the acid:
[tex]\frac{C_aV_a}{C_bV_b} = \frac{n_a}{n_b}[/tex], where
Ca = concentration of acid, Cb = concentration of base (1.0), Va = volume of acid (250 mL), Vb = volume of base (13.48 mL), na = number of moles of acid and (1) nb = number of moles of base (1).
[tex]C_a[/tex] = [tex]\frac{1 X 13.48 X 1}{250.0 X 1}[/tex]
   = 0.05392 M
Concentration = mole/volume
0.05392 = mole/0.25
mole of acid = 0.01348
Also, mole = mass/molar mass
Molar mass = mass/mole
           = 0.79/0.01348
            = 58.61 g/mol
The molar mass of the acid is 58.61 g/mol.
