This is for Gravimetric Determination of Chloride. I need help solving and fill

Question

This is for Gravimetric Determination of Chloride. I need help solving and fill up the remaining box. Can you help me here?

Trial 1

crucible mass (g)

28.1273

crucible + ppt mass (g)

29.0207

ppt mass (AgCl) (g)

mol of AgCl = mol of Cl-

mass chloride in unknown chloride sample (g)

% chloride in unknown

chloride sample

average (%)

standard deviation (%)

Trial 1

crucible mass (g)

28.1273

crucible + ppt mass (g)

29.0207

ppt mass (AgCl) (g)

mol of AgCl = mol of Cl-

mass chloride in unknown chloride sample (g)

% chloride in unknown

chloride sample

average (%)

standard deviation (%)

Explanation / Answer

Solution:

ppt mass (AgCl)(g) = [crucible + ppt mass(g) - crucible mass(g)] = [29.0207 - 28.1273] = 0.8934

MWAgCl = 143.32 g/mol; MWCl = 35.4527 g/mol.

mass chloride in unknown chloride sample(g) = 107.86 g/mol

% chloride in unknown chloride sample = 54.25% Cl by mass

average(%) = 89%

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