Answer:
the molar mass of Acetylsalicylic acid is 180.2 g.
Explanation:
Knowing that Acetylsalicylic acid (AA-H) has one acidic proton, we can write the equation for the reaction with NaOH as follows:
AA-<u>H </u>+ NaOH → AA-Na + H₂O (underlined is the acidic proton)
From the equation, we know that 1 mol NaOH reacts with 1 mol AA-H.
The moles of NaOH that react with AA-H can be calculated using the data provided by the problem since we have the volume and the concentration of the NaOH used:
mol NaOH = V * C where V = volume and C = concentration.
The molar concentration (M) is the number of moles present in 1 liter solution, then, the NaOH solution used has 0.5065 mol NaOH in 1000 ml.
The moles of NaOH used in the reaction can be calculated this way:
mol NaOH = 35.17 ml * (0.5065 mol / 1000 ml) = 0.01781 mol
Then, because 1 mol NaOH reacts with 1 mol AA-H, there must be 0.01781 mol AA-H that react with 0.01781 mol NaOH.
This number of mol of AA-H has a mass of 3.210 g according to the information provided by the problem. Then the mass of a single mol (the molar mass) must be:
molar mass = 1 mol * 3.210 g / 0.01781 mol = <u>180.2 g</u>
