A student prepares a weak acid solution by dissolving 0.3000 g HZ to give 100. mL solution. The titration requires 33.5 mL of 0.1025 M NaOH. Calculate the molar mass of the acid. g/mol (a) Would the molar mass be too high, too low, or unaffected if the student accidentally used 0.4000 g in the calculation? Explain. since the molar mass and mass of acid are The molar mass would be - elect select - ▼ (b) Would the molar mass be too high, too low, or unaffected if the student accidentally used 23.5 mL instead of 33.5 mL? Explain. The molar mass would be - Select - since the molar mass and number of moles which depends on the volume are

87.37g/mol

A. 116.49g/mol

The molar mass of HZ increased

B. 124.5459 g/mol

The molar mass of HZ increased.

Explanation:

For a neutralisation reaction,

HZ (aq) + NaOH (aq) - -> NaZ + H2O (l)

So from the above equation, 1 mole of acid, HZ reacted with 1 mole of base, NaOH to give 1 mole of salt.

Calculating moles of NaOH = molarity (molar concentration) * volume of solution

Molarity is defined as the number of moles of solute per liter of solution.

Moles of NaOH = 33.5 x 10^-3 * 0.1025

= 3.43375 x 10^-3 mol

Using stoichiometry, therefore there were 3.43375 x 10^-3 mol of HZ reacted.

Number of moles = mass/molar mass

Therefore, molar mass = 0.300/3.43375 x 10^-3 mol

= 87.37g/mol

Molar mass of a chemical compound is defined as the mass of a sample of that compound per the amount of substance in that sample in moles.

A. Molar mass = 0.400/3.43375 x 10^-3

= 116.49g/mol

The molar mass of HZ increased.

B. Moles of NaOH = 23.5 x 10^-3 * 0.1025

= 2.40875 x 10^-3 mol

By stoichiometry, therefore 2.40875 x 10^-3 mol of HZ reacted.

Number of moles = mass/molar mass

Therefore, molar mass = 0.300/2.40875 x 10^-3

= 124.5459 g/mol

The molar mass of HZ increased.