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Diketahui : BM NaCl = 58,5 ; valensi = 1 Bobot sampel 150 mg V AgNO3 total 25 mL N AgNO3 0,1 N V NH 4 CNS 0,1N = 9,5 mL AgNO3 + NaCl AgCl + NaNO3 V AgNO3sisa x N AgNO3 = V NH 4 CNS x N NH 4 CNS V AgNO3sisa x 0,1 N = 9,5 mL x 0,1N V AgNO3sisa = 9,5 mL V AgNO3bereaksi = V AgNO3total ‐ V AgNO3sisa = 25 mL – 9,5 mL = 15,5 mL Mol AgNO3 = M x V ; M AgNO3 = 0,1N/1 = 0,1 M = 0,1M x 15,5 mL = 1,55 mmol Mol NaCl = 1/1 x mol AgNO3 = 1,55 mmol Massa NaCl = mol x BM = 1,55 mmol x 58,5 = 90,675 mg Kadar NaCl dalam sampel = (90,675 mg/150 mg) x 100% = 60,45% b/b

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Diketahui:

BMNaCl=58,5;valensi=1

Bobotsampel150mg

VAgNO3total25mL

NAgNO30,1N

VNH4CNS0,1N=9,5mL

AgNO3+NaClAgCl+NaNO3VAgNO3sisaxNAgNO3=VNH4CNSxNNH4CNS

VAgNO3sisax0,1N=9,5mLx0,1N

VAgNO3sisa=9,5mL

VAgNO3bereaksi=VAgNO3totalVAgNO3sisa

=25mL9,5mL

=15,5mL

MolAgNO3=MxV ;MAgNO3=0,1N/1=0,1M

=0,1Mx15,5mL=1,55mmol

MolNaCl=1/1xmolAgNO3=1,55mmol

MassaNaCl=molxBM

=1,55mmolx58,5

=90,675mg

KadarNaCldalamsampel=(90,675mg/150mg)x100%

=60,45%b/b