Question

A student completes a titration of an unknown monoproticacid. In this experiment, 0.79 g of the acid is dissolved in 250.0 mL of water. It requires 13.48mL of 1.0 M NaOH to reach theequivalence point. What is the molar mass of the acid?

Answer 1

58.61 g/mol

Step-by-step 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:

`(C_aV_a)/(C_bV_b) = (n_a)/(n_b)`, 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).

`C_a` =
`(1 X 13.48 X 1)/(250.0 X 1)`

= 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.

Answer 2

58.6g/mol

Step-by-step 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.