Question

0.456 grams of a monoprotic acid is titrated with 45.88 mL of 0.0500 M NaOH. What is the molecular mass (molar mass) of the acid

Answer 1

`Molar\ mass= 198.78\ g/mol`

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Moles =Molarity * {Volume\ of\ the\ solution}`

Given :

For NaOH :

Molarity = 0.0500 M

Volume = 45.88 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 45.88×10⁻³ L

`Moles_(NaOH) =0.0500 * {45.88* 10^(-3)}\ moles=0.002294\ moles`

Moles of NaOH = Moles of monoprotic acid

So, moles of monoprotic acid = 0.002294 moles

Given that:- mass = 0.456 g

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`0.002294\ moles= (0.456\ g)/(Molar\ mass)`

`Molar\ mass= 198.78\ g/mol`