25 ml tartaric solution is tritrated with 31 ml ammonium equation rearranged for normality of ammonium: starting with N1V1=N2V2

N1=
V1=
v2=
final answer N2=

I answered a question almost like this a couple of days ago. I hope you aren't the same student for there is no such thing as ammonium by itself. Ammonia yes.

You don't have any concentrations of either the tartaric acid or the ammonia but here is how you do it when you get those values.
mols tartaric acid = gram/molar mass
M tartaric acid x 2 = N tartaric acid.
Then N tar x mL tar = mL NH3 x N NH3.
Solve for the unknown wherever that is.