A 0.2119 g sample of an acid, HA, was neutralized by 34.7 mL 0.0500 M NaOH solution. The molecular mass of the acid is ________.

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of NaOH used to neutralize the acid sample:

0.0500 M NaOH = 0.0500 moles of NaOH per liter of solution

34.7 mL = 0.0347 L

Amount of NaOH used = 0.0500 moles/L x 0.0347 L = 0.001735 moles of NaOH

Since the acid and the base react in a 1:1 ratio, the amount of acid in the sample is also 0.001735 moles.

Now we can use the formula for molecular mass:

Molecular mass = (mass of sample) / (amount of substance)

Molecular mass = 0.2119 g / 0.001735 mol = 122 g/mol

Therefore, the molecular mass of the acid is 122 g/mol, which corresponds to choice C.

A 02119 G Sample of an Acid, HA, Was Neutralized by by 34.7