Respuesta :
The given question is incomplete. The complete question is as follows.
An analytical chemist weighs out 0.093 g of an unknown monoprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.1600 M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 6.5 mL of NaOH solution. Calculate the molar mass of the unknown acid. Round your answer to 2 significant digits.
Explanation:
The given data is as follows.
Mass of monoprotic acid = 0.093 g,
Volume of monoprotic acid = 250 mL,
[NaOH] = 0.1600 M, and volume of NaOH required at reaches the equivalence point = 6.5 mL.
Now, first of all we will calculate the moles of NaOH as follows.
Moles of NaOH = molarity × volume (L)
= 0.1600 M × 0.0065 L
= 0.00104 moles
Let us assume that monoprotic acid is HA. And, it is known that monoprotic acid has 1 hydrogen ion and NaOH is also monoprotic base, means it has 1 hydroxide ion. Therefore, the reaction will be as follows.
[tex]HA + NaOH \rightarrow NaA + H_{2}O[/tex]
Hence, 1 mole of NaOH = 1 mole of HA
So, 0.00104 moles of NaOH = 0.00104 moles of HA
Now, we will calculate the molar mass of unknown acid as follows.
Molar mass of unknown acid = [tex]\frac{mass}{moles}[/tex]
= [tex]\frac{0.093 g}{0.00104 moles}[/tex]
= 89.42 g/mol
Thus, we can conclude that molar mass of the unknown acid is 89.42 g/mol.
