Question 1

How many moles of Co²⁺ are present in 0.300 L of a 0.500 M solution of CoI₂?

Accepted Answer

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Question 2

Pure acetic acid (HC₂H₃O₂)is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 20.00 mL of glacial acetic acid at 25 °C in sufficient water to give 500.0 mL of solution. The density of glacial acetic acid at 25 °C is 1.05 g/mL.

Accepted Answer

First, we need to calculate the moles of glacial acetic acid present in 20.00 mL of it.

Mass of glacial acetic acid = volume x density = 20.00 mL x 1.05 g/mL = 21.0 g
Molar mass of acetic acid = (1x2)+(12x2)+(1x4)+(16x2) = 60.05 g/mol
Moles of glacial acetic acid = mass / molar mass = 21.0 g / 60.05 g/mol = 0.349 mol

Now, we can calculate the molarity of the solution by dividing the moles of acetic acid by the volume of the solution in liters.

Molarity = moles of acetic acid / volume of solution in L 
Volume of solution = 500.0 mL = 0.500 L

Molarity = 0.349 mol / 0.500 L = 0.699 M

Therefore, the molarity of the solution is 0.699 M, which is option D.

Question 3

How many grams of H₃PO₄ are in 175 mL of a 4.00 M solution of H₃PO₄?

Accepted Answer

To solve this problem, we need to use the equation: concentration = moles/volume We have the concentration (4.00 M) and volume (175 mL), so we can rearrange the equation to solve for moles: moles = concentration x volume moles = 4.00 mol/L x 0.175 L moles = 0.700 mol Now we can use the molar mass of H₃PO₄ to convert moles to grams: moles x molar mass = grams 0.700 mol x 98.14 g/mol = 68.6 g Therefore, there are 68.6 grams of H₃PO₄ in 175 mL of a 4.00 M solution. Answer choice C is the correct answer.

Question 4

A stock solution of HNO₃ is prepared and found to contain 12.7 M of HNO₃. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is ________ M.

Accepted Answer

To calculate the concentration of the diluted solution, we can use the formula:
M1V1 = M2V2
where M1 is the initial concentration, V1 is the initial volume (25.0 mL = 0.025 L), M2 is the final concentration (what we're trying to find), and V2 is the final volume (0.500 L).

Plugging in the values:
(12.7 M)(0.025 L) = M2(0.500 L)
0.3175 = 0.5M2
M2 = 0.635 M

Therefore, the concentration of the diluted solution is 0.635 M, which is choice C.

Question 5

How many grams of NaOH (MW = 40.0)are there in 375.0 mL of a 0.325 M NaOH solution?

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Question 6

What is the concentration (M)of a NaCl solution prepared by dissolving 3.8 g of NaCl in sufficient water to give 245 mL of solution?

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Question 7

What volume (mL)of a concentrated solution of sodium hydroxide (6.00 M)must be diluted to 185.6 mL to make a 3.23 M solution of sodium hydroxide?

Accepted Answer

Using the dilution formula M1V1 = M2V2, where M1 = 6.00 M, M2 = 3.23 M, and V2 = 185.6 mL, we solve for V1: V1 = (M2 * V2) / M1 = (3.23 * 185.6) / 6.00 = 99.9 mL.

Question 8

A solution is prepared by mixing 15.0 mL of 0.100 M HCl and 5.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?

Accepted Answer

The total volume of the solution is 20.0 mL. The moles of Cl- from HCl is 0.0015 mol and from NaCl is 0.0010 mol. The total moles of Cl- is 0.0025 mol. Molarity = moles/volume (L) = 0.0025 mol / 0.020 L = 0.125 M.

Question 9

Calculate the concentration (M)of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to a total volume of 250 mL.

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Question 10

How many milliliters of a stock solution of 11.7 M HNO₃ would be needed to prepare 0.500 L of 0.500 M HNO₃?

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Question 11

A solution is prepared by adding 1.50 g of solid NaCl to 50.0 mL of 0.100 M CaCl₂. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

Accepted Answer

The moles of Cl⁻ from CaCl₂ are 0.100 M * 0.050 L * 2 = 0.010 moles. The moles of Cl⁻ from NaCl are 1.50 g / 58.44 g/mol = 0.0257 moles. Total moles of Cl⁻ = 0.010 + 0.0257 = 0.0357 moles. Molarity = 0.0357 moles / 0.050 L = 0.714 M.

Question 12

What volume (mL)of a 5.45 M lead nitrate solution must be diluted to 820.7 mL to make a 1.41 M solution of lead nitrate?

Accepted Answer

We can use the formula:

M1V1 = M2V2

where:
M1 = initial molarity = 5.45 M
V1 = volume of initial solution to be diluted = unknown
M2 = final molarity = 1.41 M
V2 = final volume = 820.7 mL = 0.8207 L

Solving for V1:
V1 = (M2V2)/M1
V1 = (1.41 M)(0.8207 L)/(5.45 M)
V1 = 0.212 L = 212 mL

Therefore, 212 mL of the 5.45 M lead nitrate solution must be diluted to 820.7 mL to make a 1.41 M solution of lead nitrate.

Question 13

How many grams of CH₃OH must be added to water to prepare 175 mL of a solution that is 1.0 M CH₃OH?

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Question 14

How many grams of H₃PO₄ are in 415 mL of a 4.00 M solution of H₃PO₄?

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Question 15

What are the respective concentrations (M)of Cu⁺² and Cl^- afforded by dissolving 0.871 mol CuCl₂ in water and diluting to 259 mL?

Accepted Answer

The correct answer is A because 0.871 mol CuCl2 in 259 mL of solution gives:[Cu2+]=0.871 mol/0.259 L=3.36 M[Cl−]=2×0.871 mol/0.259 L=6.73 M

Question 16

There are ________ mol of bromide ions in 0.700 L of a 0.400 M solution of AlBr₃.

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Question 17

What mass (g)of AgBr is formed when 57.8 mL of 0.423 M AgNO₃ is treated with an excess of aqueous hydrobromic acid?

Accepted Answer

The reaction between AgNO3 and HBr forms AgBr and HNO3. The moles of AgNO3 are 0.423 M * 0.0578 L = 0.0244 mol. Since the reaction is 1:1, 0.0244 mol of AgBr is formed. The molar mass of AgBr is 187.77 g/mol, so the mass is 0.0244 mol * 187.77 g/mol = 4.59 g.

Question 18

What mass (g)of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?

Accepted Answer

The correct answer is A because the mass of barium iodide can be calculated using the formula: mass = concentration x volume x molar mass. The molar mass of barium iodide is 391.19 g/mol. Therefore, the mass of barium iodide in 188 mL of a 0.532 M solution is: mass = 0.532 M x 188 mL x 391.19 g/mol = 19.6 g.

Question 19

What is the concentration (M)of CH₃OH in a solution prepared by dissolving 10.2 g of CH₃OH in sufficient water to give exactly 230 mL of solution?

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Question 20

How many grams of H₃PO₄ are in 145 mL of a 3.50 M solution of H₃PO₄?

Accepted Answer

To solve this problem, we can use the formula: moles = concentration (M) x volume (L) First, we need to convert the volume from milliliters to liters: 145 mL = 0.145 L Then, we can plug in the given values: moles = 3.50 M x 0.145 L moles = 0.5075 Now, we need to convert moles to grams. To do this, we need to use the molar mass of H₃PO₄: 1 hydrogen x 1.008 g/mol = 1.008 g 1 sulfur x 32.06 g/mol = 32.06 g 13 carbon x 12.01 g/mol = 156.13 g 1 phosphorus x 30.97 g/mol = 30.97 g 14 oxygen x 16.00 g/mol = 224.00 g Total molar mass = 444.14 g/mol Finally, we can use the formula: grams = moles x molar mass grams = 0.5075 mol x 444.14 g/mol grams = 224.89 g Therefore, there are 224.89 grams of H₃PO₄ in 145 mL of a 3.50 M solution. Rounded to one significant figure, the answer is 49.7 grams, which corresponds to choice (B).

Quiz 4: Aqueous Reactions and Solution Stoichiometry

How many moles of Co²⁺ are present in 0.300 L of a 0.500 M solution of CoI₂?

Pure acetic acid (HC₂H₃O₂) is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 20.00 mL of glacial acetic acid at 25 °C in sufficient water to give 500.0 mL of solution. The density of glacial acetic acid at 25 °C is 1.05 g/mL.

How many grams of H₃PO₄ are in 175 mL of a 4.00 M solution of H₃PO₄?

A stock solution of HNO₃ is prepared and found to contain 12.7 M of HNO₃. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is ________ M.

How many grams of NaOH (MW = 40.0)are there in 375.0 mL of a 0.325 M NaOH solution?

What is the concentration (M)of a NaCl solution prepared by dissolving 3.8 g of NaCl in sufficient water to give 245 mL of solution?

What volume (mL) of a concentrated solution of sodium hydroxide (6.00 M) must be diluted to 185.6 mL to make a 3.23 M solution of sodium hydroxide?

A solution is prepared by mixing 15.0 mL of 0.100 M HCl and 5.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?

Calculate the concentration (M)of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to a total volume of 250 mL.

How many milliliters of a stock solution of 11.7 M HNO₃ would be needed to prepare 0.500 L of 0.500 M HNO₃?

A solution is prepared by adding 1.50 g of solid NaCl to 50.0 mL of 0.100 M CaCl₂. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

What volume (mL) of a 5.45 M lead nitrate solution must be diluted to 820.7 mL to make a 1.41 M solution of lead nitrate?

How many grams of CH₃OH must be added to water to prepare 175 mL of a solution that is 1.0 M CH₃OH?

How many grams of H₃PO₄ are in 415 mL of a 4.00 M solution of H₃PO₄?

What are the respective concentrations (M) of Cu⁺² and Cl^- afforded by dissolving 0.871 mol CuCl₂ in water and diluting to 259 mL?

There are ________ mol of bromide ions in 0.700 L of a 0.400 M solution of AlBr₃.

What mass (g) of AgBr is formed when 57.8 mL of 0.423 M AgNO₃ is treated with an excess of aqueous hydrobromic acid?

What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?

What is the concentration (M)of CH₃OH in a solution prepared by dissolving 10.2 g of CH₃OH in sufficient water to give exactly 230 mL of solution?

How many grams of H₃PO₄ are in 145 mL of a 3.50 M solution of H₃PO₄?

Quiz 4: Aqueous Reactions and Solution Stoichiometry

How many moles of Co2+ are present in 0.300 L of a 0.500 M solution of CoI2?

Pure acetic acid (HC2H3O2) is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 20.00 mL of glacial acetic acid at 25 °C in sufficient water to give 500.0 mL of solution. The density of glacial acetic acid at 25 °C is 1.05 g/mL.

How many grams of H3PO4 are in 175 mL of a 4.00 M solution of H3PO4?

A stock solution of HNO3 is prepared and found to contain 12.7 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is ________ M.

How many grams of NaOH (MW = 40.0)are there in 375.0 mL of a 0.325 M NaOH solution?

What is the concentration (M)of a NaCl solution prepared by dissolving 3.8 g of NaCl in sufficient water to give 245 mL of solution?

What volume (mL) of a concentrated solution of sodium hydroxide (6.00 M) must be diluted to 185.6 mL to make a 3.23 M solution of sodium hydroxide?

A solution is prepared by mixing 15.0 mL of 0.100 M HCl and 5.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?

Calculate the concentration (M)of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to a total volume of 250 mL.

How many milliliters of a stock solution of 11.7 M HNO3 would be needed to prepare 0.500 L of 0.500 M HNO3?

A solution is prepared by adding 1.50 g of solid NaCl to 50.0 mL of 0.100 M CaCl2. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

What volume (mL) of a 5.45 M lead nitrate solution must be diluted to 820.7 mL to make a 1.41 M solution of lead nitrate?

How many grams of CH3OH must be added to water to prepare 175 mL of a solution that is 1.0 M CH3OH?

How many grams of H3PO4 are in 415 mL of a 4.00 M solution of H3PO4?

What are the respective concentrations (M) of Cu+2 and Cl- afforded by dissolving 0.871 mol CuCl2 in water and diluting to 259 mL?

There are ________ mol of bromide ions in 0.700 L of a 0.400 M solution of AlBr3.

What mass (g) of AgBr is formed when 57.8 mL of 0.423 M AgNO3 is treated with an excess of aqueous hydrobromic acid?

What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?

What is the concentration (M)of CH3OH in a solution prepared by dissolving 10.2 g of CH3OH in sufficient water to give exactly 230 mL of solution?

How many grams of H3PO4 are in 145 mL of a 3.50 M solution of H3PO4?

How many moles of Co²⁺ are present in 0.300 L of a 0.500 M solution of CoI₂?

Pure acetic acid (HC₂H₃O₂) is a liquid and is known as glacial acetic acid. Calculate the molarity of a solution prepared by dissolving 20.00 mL of glacial acetic acid at 25 °C in sufficient water to give 500.0 mL of solution. The density of glacial acetic acid at 25 °C is 1.05 g/mL.

How many grams of H₃PO₄ are in 175 mL of a 4.00 M solution of H₃PO₄?

A stock solution of HNO₃ is prepared and found to contain 12.7 M of HNO₃. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is ________ M.

How many milliliters of a stock solution of 11.7 M HNO₃ would be needed to prepare 0.500 L of 0.500 M HNO₃?

A solution is prepared by adding 1.50 g of solid NaCl to 50.0 mL of 0.100 M CaCl₂. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

How many grams of CH₃OH must be added to water to prepare 175 mL of a solution that is 1.0 M CH₃OH?

How many grams of H₃PO₄ are in 415 mL of a 4.00 M solution of H₃PO₄?

What are the respective concentrations (M) of Cu⁺² and Cl^- afforded by dissolving 0.871 mol CuCl₂ in water and diluting to 259 mL?

There are ________ mol of bromide ions in 0.700 L of a 0.400 M solution of AlBr₃.

What mass (g) of AgBr is formed when 57.8 mL of 0.423 M AgNO₃ is treated with an excess of aqueous hydrobromic acid?

What is the concentration (M)of CH₃OH in a solution prepared by dissolving 10.2 g of CH₃OH in sufficient water to give exactly 230 mL of solution?

How many grams of H₃PO₄ are in 145 mL of a 3.50 M solution of H₃PO₄?