0.103g sample of NH4NO3 required 12.8ml of 0.101M NaOH for neutralization.what is the %purity of a sample?

Answer 1

NH4NO3 + NaOH ,p==> NH4OH + NaNO3 ==> NH3 + H2O + NaNO3
millimoles NaOH = M x mL = 0.101 M x 12.8 mL = 1.29 millimoles = 0.00129 moles.
For every 1 mole NaOH you had 1 mole NH4NO3; therefore, moles NH4NO3 = 0.00129.
grams NH4NO3 = moles x molar mass = 0.00128 x 80 = 0.103 g
% purity = (grams NH4NO3/g sample)*100 = ?