A 0.629 g sample of a diprotic acid is dissolved in water and titrated with 0.270 M NaOH. What is the molar mass of the acid if 36.4 mL of the NaOH solution is required to neutralize the sample? Assume the volume of NaOH corresponds to the second equivalence point. A flask with a solution sits on the base of a ring stand. A buret filled with liquid is suspended above the flask by the ring stand. molar mass: g/mol

The molar mass of acid is 127.97 g/mol

Explanation:

To calculate the number of moles for given molarity, we use the equation:

Molarity of NaOH solution = 0.270 M

Volume of solution = 36.4 mL

Putting values in above equation, we get:

The chemical equation for the reaction of NaOH and diprotic acid follows:

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid =

To calculate the number of moles, we use the equation:

Mass of acid = 0.629 g

Moles of acid = 0.004915 moles

Putting values in above equation, we get:

Hence, the molar mass of acid is 127.97 g/mol

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.