To find the correct molarity of the original acid solution, you need to consider the volume used during the titration and the stoichiometry of the reaction.
1) Since you know that 17.8 mL of 0.10 M NaOH solution was required to reach the neutral endpoint, you can use this information to determine the number of moles of NaOH used.
0.10 M NaOH represents 0.10 moles of NaOH in 1 liter of solution, or 0.10 mmol in 1 mL of solution. Therefore, the number of moles of NaOH used in the titration is:
0.10 mmol/mL x 17.8 mL = 1.78 mmol NaOH
Since the stoichiometry of the reaction between the acid and NaOH is 1:1 (assuming the acid is monoprotic), this means that 1.78 mmol of the acid was also neutralized during the titration.
However, you need to take into account the dilution. The titrated portion was taken from the original 50.00 mL solution, so you only used half of it, which is 25.00 mL. This means that the 1.78 mmol of acid actually came from 25.00 mL of the original solution.
To find the molarity (M), you can use the formula:
Molarity = moles of solute / volume of solution (in liters)
Conversion: 25.00 mL = 25.00 mL x (1 L / 1000 mL) = 0.025 L
Molarity = 1.78 mmol / 0.025 L = 71.2 mmol/L = 0.0712 M
Hence, the molarity of the original acid solution is 0.0712 M.
Now, let's move on to the second question.
2) To find the molecular weight of the acid, you need to use the mass of the acid weighed out in grams and the number of moles of acid used during the titration.
The mass of the acid weighed out is given as 0.732 grams.
To find the number of moles (n) of the acid used, you can use the formula:
n = mass / molar mass
n = 0.732 g / molar mass
We already calculated that 44.5 mmol of acid was used during the titration. To convert this to moles:
44.5 mmol = 44.5 mmol x (1 mol / 1000 mmol) = 0.0445 mol
Now, we can set up the equation:
0.732 g / molar mass = 0.0445 mol
Rearranging the equation, we get:
molar mass = 0.732 g / 0.0445 mol
molar mass = 16.49 g/mol (rounded to two decimal places)
Therefore, the molecular weight of the acid is approximately 16.49 g/mol.