Answer to Question #341567 in Chemistry for cedrick

Question

The chloride in a sample containing approxiamately 3.5% NaCl is to be determined

gravimetrically as AgCl. What weight of sample must be taken to obtain a precipitate

that will weigh about 0.10 g?

Answer 1

Solution:

Gravimetric factor (GF):

Analyte – NaCl; Precipitate – AgCl

NaCl → AgCl

The molar mass of NaCl is 58.44 g mol⁻¹

The molar mass of AgCl is 143.32 g mol⁻¹

Therefore,

GF = (58.44 g NaCl mol⁻¹) / (143.32 g AgCl mol⁻¹) × (1/1) = 0.40776 mol NaCl / mol AgCl

GF = 0.40776

%Analyte = (Weight of product × GF × 100%) / (Weight of sample)

Therefore,

Weight of sample = (Weight of product × GF × 100%) / (%Analyte)

Weight of sample = (0.10 g × 0.40776 × 100%) / (3.5%) = 1.165 g

Weight of sample = 1.165 g

Answer: The weight of sample is 1.165 grams