Question

1) A student repeats an analysis 4 times and obtains the following calculated molarities. What is the precision of the laboratory analysis (determined as relative average deviation)? 0.5933 M, 0.5972 M, 0.5978 M, 0.5941 M

2) Find the grams of AgCl formed by the reaction of 6.41 g of ZnCl2 with 40.0ml of a 0.404 M AgNO3 solution according to the reaction : ZnCl2 (s) + 2 AgNO3 (aq) ---- Zn(NO3)2 (aq) + 2 AgCl (s)

3) How many moles of citric acid (H3C6H5O7) are present in 48.96 ml of a 0.710 M citric acid solution?

4)Calculate the grams of nitric acid (HNO3) required to prepare 375.0 ml of a 0.6500 M solution

Answer 1

1)

mean = (0.5933 + 0.5941 + 0.5972 + 0.5978 ) / 4

         = 0.5956

Average deviation = |x - | / n

                            = 0.0023 + 0.0015 +0.0016 + 0.0022 / 4

                             = 0.0019

precision = 0.5956 +/- 0.0019

2)

moles of ZnCl2 = 6.41 / 136.286 = 0.0470 mol

moles of AgNO3 = 40 x 0.404 / 1000 = 0.01616 mol

ZnCl2 (s) + 2 AgNO3 (aq) ---- Zn(NO3)2 (aq) + 2 AgCl (s)

1                     2                          1                       2

0.0470          0.0162                                      

here limiting reagent is AgNO3 .

moles of AgCl formed = 0.0162 mol

mass of AgCl formed = 0.0162 x 143.32

mass of AgCl formed = 2.32 g

Answer 2

Let's solve each question step by step:

1. Precision of Laboratory Analysis (Relative Average Deviation): To calculate the precision of the laboratory analysis, we first need to find the average molarity (M_avg) and the relative average deviation (RAD).

Average Molarity (M_avg): M_avg = (0.5933 M + 0.5972 M + 0.5978 M + 0.5941 M) / 4 M_avg = 0.5956 M

Now, calculate the deviations of each value from the average: Deviation 1 = |0.5933 M - 0.5956 M| = 0.0023 M Deviation 2 = |0.5972 M - 0.5956 M| = 0.0016 M Deviation 3 = |0.5978 M - 0.5956 M| = 0.0022 M Deviation 4 = |0.5941 M - 0.5956 M| = 0.0015 M

Now, calculate the sum of the absolute deviations: Sum of Absolute Deviations = 0.0023 M + 0.0016 M + 0.0022 M + 0.0015 M = 0.0076 M

Relative Average Deviation (RAD): RAD = (Sum of Absolute Deviations / Number of Observations) * 100 RAD = (0.0076 M / 4) * 100 RAD = 0.019 M

So, the precision of the laboratory analysis, determined as relative average deviation, is 0.019 M.

2. Grams of AgCl Formed: To find the grams of AgCl formed, we first need to calculate the moles of AgNO3 reacted and then use the stoichiometric ratio between AgNO3 and AgCl.

Moles of AgNO3: Moles of AgNO3 = Volume of AgNO3 (in liters) * Molarity of AgNO3 Moles of AgNO3 = 40.0 ml * (1 L / 1000 ml) * 0.404 M Moles of AgNO3 = 0.01616 moles

According to the balanced equation, 1 mole of ZnCl2 reacts with 2 moles of AgNO3 to form 2 moles of AgCl.

Moles of AgCl formed = 2 * Moles of AgNO3 Moles of AgCl formed = 2 * 0.01616 moles Moles of AgCl formed = 0.03232 moles

Now, calculate the molar mass of AgCl: Molar mass of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol

Grams of AgCl formed = Moles of AgCl formed * Molar mass of AgCl Grams of AgCl formed = 0.03232 moles * 143.32 g/mol Grams of AgCl formed = 4.632 g

So, 4.632 grams of AgCl are formed.

3. Moles of Citric Acid: To find the moles of citric acid (H3C6H5O7) present, we can use the molarity and volume of the solution.

Moles of Citric Acid = Molarity * Volume (in liters) Moles of Citric Acid = 0.710 M * (48.96 ml * (1 L / 1000 ml)) Moles of Citric Acid = 0.0347 moles

So, there are 0.0347 moles of citric acid present.

4. Grams of Nitric Acid Required: To calculate the grams of nitric acid required, we can use the molarity, volume, and molar mass of HNO3.

Moles of HNO3 = Molarity * Volume (in liters) Moles of HNO3 = 0.6500 M * (375.0 ml * (1 L / 1000 ml)) Moles of HNO3 = 0.2438 moles

Molar mass of HNO3 = 1.01 g/mol (H) + 14.01 g/mol (N) + 3 * 16.00 g/mol (O) = 63.01 g/mol

Grams of HNO3 required = Moles of HNO3 * Molar mass of HNO3 Grams of HNO3 required = 0.2438 moles * 63.01 g/mol Grams of HNO3 required = 15.35 g

So, 15.35 grams of nitric acid (HNO3) are required to prepare 375.0 ml of a 0.6500 M solution.

Answer 3

1. To calculate the precision of the laboratory analysis, we need to find the relative average deviation. The relative average deviation is given by the formula:

Relative Average Deviation = (Standard Deviation / Average) * 100

First, let's calculate the average of the molarities:

Average Molarity = (0.5933 M + 0.5972 M + 0.5978 M + 0.5941 M) / 4 Average Molarity = 2.3824 M / 4 Average Molarity = 0.5956 M

Next, calculate the standard deviation of the molarities:

Step 1: Find the differences between each molarity and the average molarity: (0.5933 M - 0.5956 M) = -0.0023 M (0.5972 M - 0.5956 M) = 0.0016 M (0.5978 M - 0.5956 M) = 0.0022 M (0.5941 M - 0.5956 M) = -0.0015 M

Step 2: Square each difference: (-0.0023 M)^2 = 0.00000529 M^2 (0.0016 M)^2 = 0.00000256 M^2 (0.0022 M)^2 = 0.00000484 M^2 (-0.0015 M)^2 = 0.00000225 M^2

Step 3: Find the sum of squared differences: 0.00000529 M^2 + 0.00000256 M^2 + 0.00000484 M^2 + 0.00000225 M^2 = 0.00001494 M^2

Step 4: Calculate the variance: Variance = Sum of squared differences / (Number of data points - 1) Variance = 0.00001494 M^2 / (4 - 1) Variance = 0.00001494 M^2 / 3 Variance = 0.00000498 M^2

Step 5: Calculate the standard deviation: Standard Deviation = √Variance Standard Deviation = √0.00000498 M^2 Standard Deviation = 0.002234 M

Now, calculate the relative average deviation:

Relative Average Deviation = (0.002234 M / 0.5956 M) * 100 Relative Average Deviation = 0.3757%

Therefore, the precision of the laboratory analysis (relative average deviation) is approximately 0.3757%.

2. To find the grams of AgCl formed in the reaction, we need to use stoichiometry and the given information from the balanced equation:

ZnCl2 (s) + 2 AgNO3 (aq) → Zn(NO3)2 (aq) + 2 AgCl (s)

Step 1: Calculate the moles of AgNO3 used: Moles of AgNO3 = Volume of AgNO3 solution (L) × Molarity of AgNO3 (mol/L) Moles of AgNO3 = 0.0400 L × 0.404 mol/L Moles of AgNO3 = 0.01616 mol

Step 2: Use the stoichiometric ratio to find moles of AgCl formed: 1 mol of AgNO3 produces 2 mol of AgCl Moles of AgCl = 2 × Moles of AgNO3 Moles of AgCl = 2 × 0.01616 mol Moles of AgCl = 0.03232 mol

Step 3: Calculate the molar mass of AgCl: Ag: 107.87 g/mol Cl: 35.45 g/mol Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol Molar mass of AgCl = 143.32 g/mol

Step 4: Calculate the grams of AgCl formed: Grams of AgCl = Moles of AgCl × Molar mass of AgCl Grams of AgCl = 0.03232 mol × 143.32 g/mol Grams of AgCl = 4.635 g

Therefore, 4.635 grams of AgCl are formed in the reaction.

3. To find the moles of citric acid (H3C6H5O7) in the solution, we use the given information:

Volume of citric acid solution = 48.96 mL = 0.04896 L Molarity of citric acid solution = 0.710 M

Step 1: Calculate the moles of citric acid: Moles of citric acid = Volume of solution (L) × Molarity of solution (mol/L) Moles of citric acid = 0.04896 L × 0.710 mol/L Moles of citric acid = 0.034735 mol

Therefore, there are 0.034735 moles of citric acid in the solution.

4. To calculate the grams of nitric acid (HNO3) required to prepare the solution, we use the given information:

Volume of solution = 375.0 mL = 0.3750 L Molarity of solution = 0.6500 M

Step 1: Calculate the moles of nitric acid: Moles of nitric acid = Volume of solution (L) × Molarity of solution (mol/L) Moles of nitric acid = 0.3750 L × 0.6500 mol/L Moles of nitric acid = 0.24375 mol

Step 2: Calculate the molar mass of nitric acid (HNO3): H: 1.01 g/mol N: 14.01 g/mol O: 16.00 g/mol Molar mass of HNO3 = 1.01 g/mol + 14.01 g/mol + 3(16.00 g/mol) = 63.02 g/mol

Step 3: Calculate the grams of nitric acid required: Grams of nitric acid = Moles of nitric acid × Molar mass of HNO3 Grams of nitric acid = 0.24375 mol × 63.02 g/mol Grams of nitric acid = 15.37 g

Therefore, 15.37 grams of nitric acid are required to prepare 375.0 mL of a 0.6500 M solution.