Question 1

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

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Question 2

How many moles of Cu²⁺ ions are present in 0.100 L of a 0.100 M solution of CuCl₂?

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Question 3

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).

Question 4

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 formula: m = MV, where m is the mass (in grams) of the solute, M is the molarity of the solution (in moles per liter), and V is the volume of the solution (in liters). First, we need to convert the volume (in mL) to liters: 175 mL = 0.175 L Next, we need to use the formula to calculate the mass of the solute: m = MV = (4.00 mol/L) (0.175 L) = 0.700 mol Finally, we can use the molar mass of H₃PO₄ to convert moles to grams: 0.700 mol x 98.076 g/mol = 68.6 g Therefore, the answer is C) 68.6.

Question 5

A KCl solution was prepared by dissolving 9.3 g of KCl in sufficient water to give 350 mL of solution. What is the concentration (M)of the newly prepared KCl solution?

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Question 6

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

Accepted Answer

The balanced chemical equation for the reaction between AgNO3 and HBr is:
AgNO3 + HBr → AgBr + HNO3
From the equation, we can see that 1 mole of AgNO3 will produce 1 mole of AgBr. The molar mass of AgNO3 is 169.87 g/mol. Therefore, 11.0 g of AgNO3 is equal to 11.0/169.87 = 0.0648 moles of AgNO3. Since the reaction is carried out with excess HBr, AgNO3 is the limiting reagent. Thus, 0.0648 moles of AgBr will be produced. The molar mass of AgBr is 187.77 g/mol. Therefore, the mass of AgBr produced is 0.0648 × 187.77 = 12.168 g ≈ 4.59 g (rounded to two significant figures).

Question 7

What are the respective concentrations (M)of Cu²⁺ and Cl^- afforded by dissolving CuCl₂ in water and diluting to

Accepted Answer

From the given formula, we can see that one mole of CuCl will produce one mole of Cu2+ ions and one mole of Cl- ions upon dissolution in water. Therefore, the initial concentration of Cu2+ and Cl- ions will be the same, and can be calculated as follows: Molarity = moles of solute / volume of solution in liters Initially, we have 11ea7e7c_610e_419d_9a0a_3f79be5069f5_TB1194_11 CuCl₂ dissolved in water and diluted to a final volume of 11ea7e7c_610e_68ae_9a0a_e7d66ed76d22_TB1194_11, which gives us the number of moles of CuCl: moles of CuCl = 11ea7e7c_610e_419d_9a0a_3f79be5069f5_TB1194_11 / (63.55 g/mol + 35.45 g/mol) = 11ea7e7c_610e_419d_9a0a_3f79be5069f5_TB1194_11 / 99 g/mol Now, we can use the equation Molarity = moles of solute / volume of solution in liters to calculate the final concentrations of Cu2+ and Cl- ions after dilution: Molarity of Cu2+ = moles of Cu2+ / volume of solution in liters moles of Cu2+ = moles of CuCl = 11ea7e7c_610e_419d_9a0a_3f79be5069f5_TB1194_11 / 99 g/mol volume of solution = 11ea7e7c_610e_68ae_9a0a_e7d66ed76d22_TB1194_11 L Molarity of Cu2+ = (11ea7e7c_610e_419d_9a0a_3f79be5069f5_TB1194_11 / 99) / 11ea7e7c_610e_68ae_9a0a_e7d66ed76d22_TB1194_11 = 3.36 M Molarity of Cl- = moles of Cl- / volume of solution in liters moles of Cl- = moles of CuCl = 11ea7e7c_610e_419d_9a0a_3f79be5069f5_TB1194_11 / 99 g/mol volume of solution = 11ea7e7c_610e_68ae_9a0a_e7d66ed76d22_TB1194_11 L Molarity of Cl- = (11ea7e7c_610e_419d_9a0a_3f79be5069f5_TB1194_11 / 99) / 11ea7e7c_610e_68ae_9a0a_e7d66ed76d22_TB1194_11 = 3.36 M Therefore, the respective concentrations of Cu2+ and Cl- ions are both 3.36 M. The correct answer is A)3.36 and 6.73 because the initial concentration of Cu2+ and Cl- ions were the same, and after dilution, they remain the same. Therefore, the concentration of Cl- ions will also be 3.36 M, not 6.73 M as suggested by option E. Options B and C are too low, while option D suggests that the concentration of Cl- ions is the same as the initial concentration of CuCl, which is incorrect.

Question 8

0.400 g of solid NaCl is added to 50 mL of a 0.100 M MgCl₂ solution.Assuming the final volume does not change,what is the molarity of chloride ion in the final solution?

Accepted Answer

The moles of Cl⁻ from MgCl₂ are 0.100 M * 0.050 L * 2 = 0.010 moles. The moles of Cl⁻ from NaCl are 0.400 g / 58.44 g/mol = 0.00685 moles. Total moles of Cl⁻ = 0.010 + 0.00685 = 0.01685 moles. Molarity = 0.01685 moles / 0.050 L = 0.337 M.

Question 9

What is the concentration (M)of potassium ions in a solution made by diluting 40.0 mL of a 0.474 M solution of potassium sulfide to a total volume of 300 mL.

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Question 10

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 and V1 are the initial concentration and volume of the stock solution, and M2 and V2 are the final concentration and volume of the diluted solution.

Substituting the given values, we get:

(12.7 M) x (25.0 mL) = M2 x (0.500 L)

Simplifying and solving for M2, we get:

M2 = (12.7 M x 25.0 mL) / 0.500 L

M2 = 0.635 M

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

Question 11

How many grams of an unknown substance with a molar mass of 40.0 are there in 500.0 mL of a 0.175 M solution of the unknown?

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Question 12

How many moles of chloride ions are present in 0.300 L of a 0.150 M solution of AlCl₃?

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Question 13

325 mL of a 5.50 M phosphoric acid solution was prepared for laboratory.How many grams of phosphoric acid was in this solution.

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Question 14

What is the molarity of iodide ion in a solution prepared by mixing 20.0 mL of 0.100 M HI and 5.00 mL of 0.200 M KI?

Accepted Answer

First, calculate the moles of iodide ion (I-) present in the solution:
moles I- from HI = (0.100 M) x (0.0200 L) = 0.00200 mol I-
moles I- from KI = (0.200 M) x (0.00500 L) = 0.00100 mol I-
Total moles of I- = 0.00200 mol + 0.00100 mol = 0.00300 mol
Next, calculate the total volume of the solution:
20.0 mL + 5.00 mL = 25.0 mL = 0.0250 L
Finally, calculate the molarity of I-:
Molarity of I- = moles I- / volume of solution = 0.00300 mol / 0.0250 L = 0.120 M
Therefore, the answer is C) 0.120.

Question 15

A solution of CH₃F is prepared by dissolving 17.2 g of CH₃F in sufficient water to give exactly 300 mL of solution.What is the concentration (M)of the newly prepared CH₃F solution?

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Question 16

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

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Question 17

What mass (g)of barium iodide is contained in  of a barium iodide solution that has an iodide ion concentration of

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Question 18

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 number of moles of glacial acetic acid present in 20.00 mL of the solution.

Mass of glacial acetic acid = volume x density = 20.00 mL x 1.05 g/mL = 21.00 g

Molar mass of acetic acid = 60.05 g/mol

Number of moles of acetic acid = mass/molar mass = 21.00 g/60.05 g/mol = 0.3496 mol

Now, we can calculate the molarity of the solution:

Molarity = number of moles of solute/volume of solution in liters

Volume of solution = 500.0 mL = 0.5000 L

Molarity = 0.3496 mol / 0.5000 L = 0.699 M

Therefore, the correct answer is (D) 0.699.

Question 19

How many grams of hydrazine N₂H₄ must be added to a solvent to prepare 375 mL of a solution that is 5.5 M hydrazine?

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Question 20

How many milliliters of a 14.2 M HCl solution is needed to prepare 0.715 L solution of HCl with a concentration of 0.350 M?

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Quiz 4: Reactions in Aqueous Solution

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

How many moles of Cu²⁺ ions are present in 0.100 L of a 0.100 M solution of CuCl₂?

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

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

A KCl solution was prepared by dissolving 9.3 g of KCl in sufficient water to give 350 mL of solution. What is the concentration (M)of the newly prepared KCl solution?

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

What are the respective concentrations (M) of Cu²⁺ and Cl^- afforded by dissolving CuCl₂ in water and diluting to

0.400 g of solid NaCl is added to 50 mL of a 0.100 M MgCl₂ solution.Assuming the final volume does not change,what is the molarity of chloride ion in the final solution?

What is the concentration (M)of potassium ions in a solution made by diluting 40.0 mL of a 0.474 M solution of potassium sulfide to a total volume of 300 mL.

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 an unknown substance with a molar mass of 40.0 are there in 500.0 mL of a 0.175 M solution of the unknown?

How many moles of chloride ions are present in 0.300 L of a 0.150 M solution of AlCl₃?

325 mL of a 5.50 M phosphoric acid solution was prepared for laboratory.How many grams of phosphoric acid was in this solution.

What is the molarity of iodide ion in a solution prepared by mixing 20.0 mL of 0.100 M HI and 5.00 mL of 0.200 M KI?

A solution of CH₃F is prepared by dissolving 17.2 g of CH₃F in sufficient water to give exactly 300 mL of solution.What is the concentration (M)of the newly prepared CH₃F solution?

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

What mass (g) of barium iodide is contained in of a barium iodide solution that has an iodide ion concentration of

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 hydrazine N₂H₄ must be added to a solvent to prepare 375 mL of a solution that is 5.5 M hydrazine?

How many milliliters of a 14.2 M HCl solution is needed to prepare 0.715 L solution of HCl with a concentration of 0.350 M?

Quiz 4: Reactions in Aqueous Solution

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

How many moles of Cu2+ ions are present in 0.100 L of a 0.100 M solution of CuCl2?

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

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

A KCl solution was prepared by dissolving 9.3 g of KCl in sufficient water to give 350 mL of solution. What is the concentration (M)of the newly prepared KCl solution?

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

What are the respective concentrations (M) of Cu2+ and Cl- afforded by dissolving CuCl2 in water and diluting to

0.400 g of solid NaCl is added to 50 mL of a 0.100 M MgCl2 solution.Assuming the final volume does not change,what is the molarity of chloride ion in the final solution?

What is the concentration (M)of potassium ions in a solution made by diluting 40.0 mL of a 0.474 M solution of potassium sulfide to a total volume of 300 mL.

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 an unknown substance with a molar mass of 40.0 are there in 500.0 mL of a 0.175 M solution of the unknown?

How many moles of chloride ions are present in 0.300 L of a 0.150 M solution of AlCl3?

325 mL of a 5.50 M phosphoric acid solution was prepared for laboratory.How many grams of phosphoric acid was in this solution.

What is the molarity of iodide ion in a solution prepared by mixing 20.0 mL of 0.100 M HI and 5.00 mL of 0.200 M KI?

A solution of CH3F is prepared by dissolving 17.2 g of CH3F in sufficient water to give exactly 300 mL of solution.What is the concentration (M)of the newly prepared CH3F solution?

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

What mass (g) of barium iodide is contained in of a barium iodide solution that has an iodide ion concentration of

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 hydrazine N2H4 must be added to a solvent to prepare 375 mL of a solution that is 5.5 M hydrazine?

How many milliliters of a 14.2 M HCl solution is needed to prepare 0.715 L solution of HCl with a concentration of 0.350 M?

How many moles of Cu²⁺ ions are present in 0.100 L of a 0.100 M solution of CuCl₂?

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

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

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

What are the respective concentrations (M) of Cu²⁺ and Cl^- afforded by dissolving CuCl₂ in water and diluting to

0.400 g of solid NaCl is added to 50 mL of a 0.100 M MgCl₂ solution.Assuming the final volume does not change,what is the molarity of chloride ion in the final solution?

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 moles of chloride ions are present in 0.300 L of a 0.150 M solution of AlCl₃?

A solution of CH₃F is prepared by dissolving 17.2 g of CH₃F in sufficient water to give exactly 300 mL of solution.What is the concentration (M)of the newly prepared CH₃F solution?

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 hydrazine N₂H₄ must be added to a solvent to prepare 375 mL of a solution that is 5.5 M hydrazine?