Hey so we carried out this experiment:

1. We made a standard solution using 8mL of NH3 and then made it up to 250 mL in a volumetric flask. The ammonia was taken from a bottle that contained 1L of the ammonia solution.
2. We then pipetted 20 mL aliquots from the volumetric flask.
3. Titrated it against HCl of concentration 0.013408 mol/L.
4. The MEAN TITRE was 5.70 mL to reach the end point.

So now we have to find the mass of ammonia in the original solution. I've done something VERY wrong, as my answer does NOT correlate even closely with the manufacturer's claim. Help would REALLY be appreciated.

Please can you check my calculations:
NH3 (aq) + HCl (aq) --> NH4Cl
n (HCl) = n (NH3)
n = c . v
n = 0.0134 ∙ 0.0057
therefore, n (NH3) = 0.000076 M
this is the number of moles present in the 20 ml aliquot.
the amount present in the 250 mL flask is

n (NH3) in flask = 0.000076 M x (250/20)
n (NH3) in flask = 0.000955 M

then, this is where I'm unsure. I multiplied it by 250/8 because of the original 8mL ammonia dissolved.
n (NH3) = 0.000955 M x (250/8)
n (NH3) = 0.029854 M
m (NH3) = 0.029854 M x 17
m (NH3) = 0.5075 g

now, to find the original mass in the 1L
m (NH3) = 0.5075 g x (1000/8)
m (NH3) = 63.44 g

This is a really important assignment. I'll tell you the manufacturer's claim if someone gets close.