# Answer to Question #21605 in Organic Chemistry for Sadaam Ahmed

Question #21605
A sample of a substance known to contain chloride ions was dissolved in distilled water in a 1L volumetric flask. Then 25.00 mL of this solution was treated with excess silver nitrate, AgNO3 (aq). The precipitate of silver chloride, AgCl (s) was filtered and dried. The mass of the dry precipitate was 0.765 g.

a. Calculate the concentration of chloride ions.
b. If the original substance was sodium chloride, NaCl (s), what mass of it was dissolved in the volumetric flask?
1
Expert's answer
2013-01-14T10:37:20-0500
AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3
n(AgCl) = 0.765 g / M(AgCl)
M(AgCl) = 108 + 35.5 = 143.5 g/mol
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol
n(AgCl) = chloride ions = 0.00533 mol
C(chloride ions) = n/V
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)
m(NaCl)=n(NaCl)*M(NaCl)
n(NaCl) = 0.00533 mol
M(NaCl) = 23 + 35.5 = 58.5 g/mol
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)

Answer:
a) 0.00533 mol/ L
b) 0.3115 g

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#### Comments

hanson
09.07.20, 05:33

This is incorrect. Part A requires you to use 0.025L for the volume in the formula c=n/v. You used 1L. Also, part B is incorrect. You needed to use the concentration of chloride ion from part A which was 0.214mol/L and get the moles of NaCl by dividing 1L to get 0.214 moles of NaCl. Then you can multiply this with the molar mass of NaCl. So your answer would be 12.5grams

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