Question Step 1: Obtain unknown, in a 250 ml Beaker add 5 g of unknown chloride

Question

Question Step 1: Obtain unknown, in a 250 ml Beaker add 5 g of unknown chloride XCl.

Step 2: Add water to XCl, fill beaker to 100ml level by adding water, stir solution until XCl is completely dissolved.

Step 3: Add Concentrated Nitric Acid, add 1 ml of concentrated acid to XCl solution.

Step 4: Add 1M AgNO3: Obtain a 100ml graduated cylinder and fill with 1M AgNO3, add AgNO3 to XCl solution in 5 to 25 ml increments. A precipitate of AgCl will form and gradually settle. Continue to add AgNO3 until no more precipitate forms. To verify that the reaction is complete check the chemical properties of the beaker (by double-clicking on it) and confirm that all of the XCl (in solution) has been consumed.

Step 5: Filter and weigh AgCl : Obtain a 250ml Erlenmeyer flask and add a filter, pour the contents of the beaker into the flask. Remove the filter from the Erlenmeyer flask (by again selecting the filter menu or button) and save the solid contents in a watch glass. Weigh the sample and record the result.

* *Note that in an actual lab the AgCl filtered precipitate would need to be dried to remove excess water, however in this simulation the filtered precipitate is free of water.

Observations:

Q1 - Initial unknown chloride sample weight (g):

Q2 - Weight of AgCl precipitate:

Q3 - Approximate volume of AgNO3 solution added (ml):

Q4 - Moles of AgCl = ( AgCl precipitate mass/143.32 ):

Q5 - Weight of Cl- in XCl sample = (moles of AgCl)x(atomic weight of Cl- = 35.4527) :

Q6 - % Cl- in sample = (weight of Cl- in XCl sample / weight of XCl sample) x 100 % :

Q7 - What happened as excess AgNO3 solution was added to the XCl solution ?

Explanation / Answer

Q1) 5 g

Q2)1 M silver nitrate contains 108 g of silver ion and it can precipitate 35.5 g of chloride ion completely.

100ml of 1M silver nitrate = 10.8 g of silver ion. and it can precipitate 3.55 g of chloride ion

so weight of AgCl precipitate = 14.35 g

Q3) 100 ML.

Q4) moles of AgCl = 14.35/143.22 = 0.1

Q5) 0.1* 35.4527 = 3.54527 g

Q6) % OF chloride = 3.54527)* (100)/5 = 71%

q7) as there is completely chloride is precipitated no change occurs further on addition of excess silver nitrate.

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