anda di minta menyiapkan 125,0 mL AgNO3 0,0321 M. berapa gram diperlukan dari sampel yang di ketahui mengandung 99,81% AgNO3 Berdasar massa

% AgNO3 = 99,81%
[AgNO3] = 0,0321 M
V = 125 mL = 0,125 L

n AgNO3 = M × V
= 0,0321 M × 0,125 L
= 0,0040125 mol

Mr AgNO3 = Ar Ag + Ar N + 3 Ar O
= 108 + 14 + 3(16)
= 170

m AgNO3 = Mr × n
= 170 × 0,0040125 mol
= 0,682125 gram

% AgNO3 = (m AgNO3 / m sampel) × 100%
m sampel = (100% / % AgNO3) × m AgNO3
m sampel = (100% / 99,81%) × 0,682125 gram
m sampel = 0,6834235 gram