3. ANSWERS MUST BE REPORTED IN THE CORRECT NUMBER OF SIGNIFICANT FIGURES. a. Pre

Question

3. ANSWERS MUST BE REPORTED IN THE CORRECT NUMBER OF SIGNIFICANT FIGURES. a. Preparation of Titrant. Enough titrant for three standardizations and a total of three an must be made. How many grams of silver nitrate must be weighed out in order to make 750 milliliters of 0.1 N titrant? b. Standardization of the Titrant. Three samples of pure sodium chloride have been dried and dissolved in 50 ml of water containing potassium dichromate indicator (0.2 ml of 1 M solution) Titration data for the silver nitrate solution you made is:200 Run Mass NaCl, mg Volume AgNO, Solution, ml 243.3 257.2 223.6 42.89 45.37 39.48 What is the concentration of the silver nitrate solution and error expressed in standard deviation uni . A 25.00 ml aliquot of a solution prepared by dissolving 1.2324 grams of a soluble chloride in 250.0 mlwater was titrated with 0.1103 N AgNO3 and using potassium chromate as an indicator. From the ana aver is data given below, what is the percent by weight of chloride in the sample? (Report

Explanation / Answer

1.For the particular case of Ag which has a charge of +1 in the salt the Normality is equal to the Molarity so 0.1N is equal to 0.1 M

M = moles / Volume

Volume is 750 ml or 0.75 L so, we can calculate the number of moles

0.1 = moles / 0.75 L

moles = 0.75 * 0.1 = 0.075 moles

we need the molecular weight of silver nitrate which is 170 g/gmol

moles = mass / mol. weight

mass = moles * molecular weight = 0.075 * 170 = 12.75 grams of silver nitrate are required.

2.

AgNO3 + NaCl === AgCl + NaNO3

first let´s get mol. weight of NaCl which is 58 g/gmol

calculate number of moles for each trial

1: 243.3mg / 58 = 4.194 milimoles or 0.004194 moles

According to the reaction 1 mole of NaCl reacts with 1 mole of AgCl so we have 4.194 moles of AgCl, let´s calculate molarity.

Molarity = 0.004194 / 0.04289 = 0.0978 M

If you repeat this procedure for trial 2 and 3 you get

1 . 0.0978 N

2. 0.0977 N

3. 0.0976 N

Standard deviation : 7.82033 x10 -5

average value 0.09773

%of standard deviation = 7.82033 x10 -5 / 0.09773 = 0.0008 * 100 = 0.08%

The concentration is 0.09773 +/- 0.08%

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