The standardized NaOH solution was determined to be 120M whereas the mass of the sample of impure tartaric acid was 1.35g. If the initial reading of the buret volume changed from 48 ml to a final reading of 17 ml at the endpoint of the titration of the impure sample calculate the percent of pure tartaric acid in the unknown mixture.

To calculate the percent of pure tartaric acid in the unknown mixture, we need to know the stoichiometry of the reaction between tartaric acid and sodium hydroxide. Let's assume it is a 1:1 mole ratio.

First, calculate the number of moles of NaOH used in the titration:
Initial volume of NaOH = 48 mL
Final volume of NaOH = 17 mL
Volume of NaOH used = Initial volume - Final volume = 48 mL - 17 mL = 31 mL = 0.031 L (convert from mL to L)

Molarity of NaOH (M1) = 120 M
Number of moles of NaOH (n) = Molarity × volume (in L) = 120 M × 0.031 L = 3.72 moles

Since the stoichiometry is 1:1, the number of moles of tartaric acid (n) is also 3.72 moles.

Next, calculate the molecular weight of tartaric acid. The molecular weight of tartaric acid (C4H6O6) is:
C = 12.01 g/mol × 4 = 48.04 g/mol
H = 1.008 g/mol × 6 = 6.048 g/mol
O = 16 g/mol × 6 = 96 g/mol
Total molecular weight of tartaric acid = 48.04 g/mol + 6.048 g/mol + 96 g/mol = 150.09 g/mol

Now, calculate the mass of pure tartaric acid in the sample:
Mass of impure tartaric acid sample = 1.35 g
Number of moles of tartaric acid = 3.72 moles
Mass of pure tartaric acid = Number of moles × Molecular weight = 3.72 moles × 150.09 g/mol = 556.6948 g

Finally, calculate the percent of pure tartaric acid in the unknown mixture:
Percent purity of tartaric acid = (Mass of pure tartaric acid / Mass of impure tartaric acid sample) × 100%
= (556.6948 g / 1.35 g) × 100%
= 412.36%

Therefore, the percent of pure tartaric acid in the unknown mixture is approximately 412.36%.