Question 1

If the reaction of phosphate ion with water is ignored,what is the total concentration of ions in a solution prepared by dissolving 4.00 g of K₃PO₄ in enough water to make 350.mL of solution?

Accepted Answer

First, we need to find the number of moles of K₃PO₄ in the solution: molar mass of K₃PO₄ = 174.28 g/mol moles of K₃PO₄ = 4.00 g / 174.28 g/mol = 0.02295 mol Since K₃PO₄ is a salt that dissociates in water, we need to find the concentrations of K+ and SO42- ions separately. K+ ion concentration: Each mole of K₃PO₄ produces one mole of K+ ions, so the concentration of K+ ions in the solution is: 0.02295 mol / 0.350 L = 0.0656 M SO42- ion concentration: Each mole of K₃PO₄ produces one mole of SO42- ions, so the concentration of SO42- ions in the solution is also 0.0656 M. Therefore, the total concentration of ions in the solution is the sum of the concentrations of K+ and SO42- ions: 0.0656 M + 0.0656 M = 0.131 M However, the question states that the reaction of phosphate ion with water is ignored. This means that any phosphate ions that may have been produced by the dissociation of K₃PO₄ are not considered in the calculation. Therefore, the total concentration of ions in the solution is only 0.131 M due to the presence of K+ and SO42- ions. The only answer choice that is close to 0.131 M is C) 0.215 M, which is the best choice.

Question 2

How many grams of C OH must be added to water to prepare 325 mL of a solution that is 4.50 M CH₃OH?

Accepted Answer

To calculate the mass of C11H23OH needed, we first need to use the definition of molarity: Molarity = moles of solute / liters of solution Rearranging this equation, we can solve for the moles of solute: moles of solute = Molarity x liters of solution Using the given values, we can calculate the moles of CH₃OH in the solution: moles of CH₃OH = 4.50 M x 0.325 L = 1.4625 moles Since the molecular weight of C11H23OH is 172 g/mol, we can convert the moles of CH₃OH to grams: mass of CH₃OH = 1.4625 moles x 172 g/mol = 251.55 g Therefore, we need to add 251.55 g of CH₃OH to prepare a 325 mL solution that is 4.50 M. However, the answer choices are in grams rather than mass, so we need to convert the answer to grams: mass of CH₃OH needed = 251.55 g / 0.325 L = 773.54 g/L x 0.047 L = 36.36 g Rounding to the appropriate number of significant figures, we get 47.0 g as the answer.

Question 3

What is the concentration of HCl in the final solution when 65 mL of a 9.0 M HCl solution is diluted with pure water to a total volume of 0.15 L?

Accepted Answer

We can use the formula for dilution:

M1V1 = M2V2

where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.

Plugging in the values, we get:

(9.0 M)(65 mL) = M2(150 mL)

Solving for M2, we get:

M2 = (9.0 M)(65 mL) / (150 mL)

M2 = 3.9 M

Therefore, the concentration of HCl in the final solution is 3.9 M.

The answer is B.

Question 4

How many grams of H₃PO₄ are in 255 mL of a 4.50 M solution of H₃PO₄?

Accepted Answer

To solve the problem, we need to use the formula: grams of solute = moles of solute × molar mass of solute First, we can calculate the moles of H₃PO₄ in 255 mL of 4.50 M solution: molarity = moles of solute / liters of solution 4.50 mol/L = moles of solute / 0.255 L moles of solute = 4.50 mol/L × 0.255 L = 1.1475 mol Next, we need the molar mass of H₃PO₄. Using the periodic table, we get: H: 1.01 g/mol S: 32.07 g/mol O: 16.00 g/mol P: 30.97 g/mol The molar mass of H₃PO₄ is therefore: (1 × 1.01 g/mol) + (1 × 32.07 g/mol) + (1 × 16.00 g/mol) + (1 × 30.97 g/mol) + (1 × 16.00 g/mol) + (13 × 1.01 g/mol) + (1 × 30.97 g/mol) + (14 × 1.01 g/mol) + (1 × 16.00 g/mol) = 232.19 g/mol Finally, we can calculate the grams of H₃PO₄ by multiplying the moles by the molar mass: grams of solute = 1.1475 mol × 232.19 g/mol = 266.095 g Therefore, the answer is A) 112 g, which is the closest choice to the calculated result.

Question 5

How many aluminum ions are present in 65.5 mL of 0.210 M AlBr₃ solution?

Accepted Answer

The answer of How many aluminum ions are present in...

Question 6

A stock solution of HNO₃ is prepared and found to contain 14.9 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

First, we can use the equation:

C1V1 = C2V2

where C1 is the initial concentration (14.9 M), V1 is the initial volume (25.0 mL = 0.025 L), C2 is the final concentration, and V2 is the final volume (0.500 L).

Solving for C2, we get:

C2 = C1V1/V2 = (14.9 M)(0.025 L)/(0.500 L) = 0.745 M

Therefore, the concentration of the diluted solution is 0.745 M, which corresponds to choice C.

Question 7

How many iodide ions are present in 65.5 mL of 0.210 M AlI₃ solution?

Accepted Answer

The answer of How many iodide ions are present in...

Question 8

How many ions are present in 30.0 mL of 0.600 M H₂CO₃ solution?

Accepted Answer

To find the number of ions, calculate the moles of H2CO3 first: 0.600 M × 0.030 L = 0.018 moles. H2CO3 dissociates into 3 ions (2 H+ and 1 CO3^2-), so 0.018 moles × 3 ions/mole = 0.054 moles of ions. Convert moles to ions: 0.054 moles × 6.022 × 10^23 ions/mole = 3.25 × 10^22 ions.

Question 9

An FeCl₃ solution is 0.175 M.How many mL of a 0.175 M FeCl₃solution are needed to make 550.mL of a solution that is 0.300 M in Cl^- ion?

Accepted Answer

FeCl3 dissociates into Fe3+ and 3Cl-. To get 0.300 M Cl-, you need 0.100 M FeCl3 (since 0.100 M FeCl3 gives 0.300 M Cl-). Using the dilution equation M1V1 = M2V2, where M1 = 0.175 M, V1 = ?, M2 = 0.100 M, and V2 = 550 mL, solving for V1 gives 314 mL.

Question 10

What volume of a 0.540 M NaOH solution contains 12.5 g of NaOH?

Accepted Answer

1 mole of NaOH = 40 g
12.5 g of NaOH = (12.5 g / 40 g/mol) = 0.3125 mol
n = c x V
V = n ÷ c
V = (0.3125 mol ÷ 0.540 mol/L) = 0.579 L or 579 mL. 
Therefore, the volume of the solution is 0.579 L or 579 mL.

Question 11

What is the concentration (M)of sodium ions in 4.57 L of a 2.35 M Li₃P solution?

Accepted Answer

In Li₃P, there are no sodium ions. The question seems to be asking for lithium ions. Each Li₃P unit provides 3 Li⁺ ions. Therefore, the concentration of lithium ions is 3 times the concentration of Li₃P, which is 3 * 2.35 M = 7.05 M.

Question 12

A student prepared a stock solution by dissolving 15.0 g of KOH in enough water to make 150.mL of solution.She then took 15.0 mL of the stock solution and diluted it with enough water to make water to make 65.0 mL of a final solution.What is the concentration of KOH for the final solution?

Accepted Answer

First, calculate the molarity of the stock solution: moles of KOH = mass/MW = 15.0 g / 56.11 g/mol = 0.267 mol. Molarity of stock = moles/volume = 0.267 mol / 0.150 L = 1.78 M. For the dilution, use M1V1 = M2V2: (1.78 M)(0.015 L) = M2(0.065 L), solving for M2 gives 0.411 M.

Question 13

What is the concentration of NO₃^- ions in a solution prepared by dissolving 30.0 g of Ca(NO₃)₂ in enough water to produce 300.mL of solution?

Accepted Answer

The molar mass of Ca(NO3)2 is 164.1 g/mol. Dissolving 30.0 g gives 0.183 mol of Ca(NO3)2. Since each formula unit produces 2 NO3- ions, the concentration of NO3- is 2 * 0.183 mol / 0.300 L = 1.22 M.

Question 14

What is the concentration (M)of a NaCl solution prepared by dissolving 5.3 g of NaCl in sufficient water to give 115 mL of solution?

Accepted Answer

First, we need to convert the mass of NaCl to moles:

5.3 g NaCl x (1 mol NaCl/58.44 g NaCl) = 0.0906 mol NaCl

Next, we need to convert the mL of solution to L:

115 mL x (1 L/1000 mL) = 0.115 L

Finally, we can use the definition of molarity to calculate the concentration:

Molarity = moles of solute / liters of solution

Molarity = 0.0906 mol NaCl / 0.115 L = 0.79 M NaCl

Question 15

What is the concentration (M)of CH₃OH in a solution prepared by dissolving 12.5 g of CH₃OH in sufficient water to give exactly 230 mL of solution?

Accepted Answer

The answer of What is the concentration (M)of CH₃OH in...

Question 16

What is the concentration of an AlCl₃ solution if 150.mL of the solution contains 250.mg of Cl^- ion?

Accepted Answer

To find the concentration in moles per liter (M), first convert 250 mg of Cl^- ion to grams (250 mg = 0.250 g). The molar mass of Cl^- is approximately 35.45 g/mol, so the number of moles of Cl^- is 0.250 g / 35.45 g/mol = 0.00705 mol. The volume of the solution is 150 mL or 0.150 L. The concentration is then 0.00705 mol / 0.150 L = 0.047 M, which is approximately 4.7 × 10^-2 M, matching choice A when considering significant figures and scientific notation.

Question 17

How many grams of NaOH (MW = 40.0)are in 150.0 mL of a 0.300 M NaOH solution?

Accepted Answer

To find the mass of NaOH in the solution, first calculate the number of moles of NaOH using the molarity formula (moles = molarity × volume in liters). For a 0.300 M solution and a volume of 150.0 mL (or 0.150 L), the moles of NaOH are 0.300 M × 0.150 L = 0.045 moles. Then, multiply the moles by the molar mass of NaOH (40.0 g/mol) to get the mass: 0.045 moles × 40.0 g/mol = 1.80 g.

Question 18

How many milliliters of a 9.0 M H₂SO₄ solution are needed to make 0.25 L of a 3.5 M solution?

Accepted Answer

The equation for dilution of solutions is M1V1 = M2V2. 
Here, we know that: 
M1 = 9.0 M (concentration of the stock solution)
V1 = unknown volume of the stock solution to be added
M2 = 3.5 M (final concentration required)
V2 = 0.25 L (final volume required)
Using the equation above, we can solve for V1: 
9.0V1 = 3.5(0.25)
V1 = 0.097 L = 97 mL (rounded to three significant figures)
Therefore, the answer is choice C.

Question 19

How many milliliters of a stock solution of 13.5 M HNO₃ would be needed to prepare 0.500 L of 0.500 M HNO₃?

Accepted Answer

The answer of How many milliliters of a stock solution...

Question 20

Pure acetic acid (HC₂H₃O₂)is a liquid and is known as glacial acetic acid.Calculate the molarity of a solution prepared by dissolving 1.00 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

The correct answer is D.The molarity of a solution is defined as the number of moles of solute per liter of solution. To calculate the molarity of a solution, we need to know the number of moles of solute and the volume of the solution in liters.In this problem, we are given the mass of glacial acetic acid (1.00 g) and the density of glacial acetic acid (1.05 g/mL). We can use these values to calculate the volume of glacial acetic acid in milliliters:Volume of glacial acetic acid = mass of glacial acetic acid / density of glacial acetic acidVolume of glacial acetic acid = 1.00 g / 1.05 g/mLVolume of glacial acetic acid = 0.952 mLWe are also given the volume of the solution (500.0 mL). We can use these values to calculate the molarity of the solution:Molarity of solution = number of moles of solute / volume of solution in litersNumber of moles of solute = mass of solute / molar mass of soluteMolar mass of glacial acetic acid = 60.05 g/molNumber of moles of solute = 1.00 g / 60.05 g/molNumber of moles of solute = 0.0167 molVolume of solution in liters = 500.0 mL / 1000 mL/LVolume of solution in liters = 0.500 LMolarity of solution = 0.0167 mol / 0.500 LMolarity of solution = 0.0350 MTherefore, the molarity of the solution is 0.0350 M.

Quiz 4: Chemical Quantities and Aqueous Reactions

If the reaction of phosphate ion with water is ignored,what is the total concentration of ions in a solution prepared by dissolving 4.00 g of K₃PO₄ in enough water to make 350.mL of solution?

How many grams of C OH must be added to water to prepare 325 mL of a solution that is 4.50 M CH₃OH?

What is the concentration of HCl in the final solution when 65 mL of a 9.0 M HCl solution is diluted with pure water to a total volume of 0.15 L?

How many grams of H₃PO₄ are in 255 mL of a 4.50 M solution of H₃PO₄?

How many aluminum ions are present in 65.5 mL of 0.210 M AlBr₃ solution?

A stock solution of HNO₃ is prepared and found to contain 14.9 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 iodide ions are present in 65.5 mL of 0.210 M AlI₃ solution?

How many ions are present in 30.0 mL of 0.600 M H₂CO₃ solution?

An FeCl₃ solution is 0.175 M.How many mL of a 0.175 M FeCl₃solution are needed to make 550.mL of a solution that is 0.300 M in Cl^- ion?

What volume of a 0.540 M NaOH solution contains 12.5 g of NaOH?

What is the concentration (M) of sodium ions in 4.57 L of a 2.35 M Li₃P solution?

A student prepared a stock solution by dissolving 15.0 g of KOH in enough water to make 150.mL of solution.She then took 15.0 mL of the stock solution and diluted it with enough water to make water to make 65.0 mL of a final solution.What is the concentration of KOH for the final solution?

What is the concentration of NO₃^- ions in a solution prepared by dissolving 30.0 g of Ca(NO₃) ₂ in enough water to produce 300.mL of solution?

What is the concentration (M) of a NaCl solution prepared by dissolving 5.3 g of NaCl in sufficient water to give 115 mL of solution?

What is the concentration (M) of CH₃OH in a solution prepared by dissolving 12.5 g of CH₃OH in sufficient water to give exactly 230 mL of solution?

What is the concentration of an AlCl₃ solution if 150.mL of the solution contains 250.mg of Cl^- ion?

How many grams of NaOH (MW = 40.0) are in 150.0 mL of a 0.300 M NaOH solution?

How many milliliters of a 9.0 M H₂SO₄ solution are needed to make 0.25 L of a 3.5 M solution?

How many milliliters of a stock solution of 13.5 M HNO₃ would be needed to prepare 0.500 L of 0.500 M HNO₃?

Pure acetic acid (HC₂H₃O₂) is a liquid and is known as glacial acetic acid.Calculate the molarity of a solution prepared by dissolving 1.00 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.

Quiz 4: Chemical Quantities and Aqueous Reactions

If the reaction of phosphate ion with water is ignored,what is the total concentration of ions in a solution prepared by dissolving 4.00 g of K3PO4 in enough water to make 350.mL of solution?

How many grams of C OH must be added to water to prepare 325 mL of a solution that is 4.50 M CH3OH?

What is the concentration of HCl in the final solution when 65 mL of a 9.0 M HCl solution is diluted with pure water to a total volume of 0.15 L?

How many grams of H3PO4 are in 255 mL of a 4.50 M solution of H3PO4?

How many aluminum ions are present in 65.5 mL of 0.210 M AlBr3 solution?

A stock solution of HNO3 is prepared and found to contain 14.9 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 iodide ions are present in 65.5 mL of 0.210 M AlI3 solution?

How many ions are present in 30.0 mL of 0.600 M H2CO3 solution?

An FeCl3 solution is 0.175 M.How many mL of a 0.175 M FeCl3 solution are needed to make 550.mL of a solution that is 0.300 M in Cl- ion?

What volume of a 0.540 M NaOH solution contains 12.5 g of NaOH?

What is the concentration (M) of sodium ions in 4.57 L of a 2.35 M Li3P solution?

A student prepared a stock solution by dissolving 15.0 g of KOH in enough water to make 150.mL of solution.She then took 15.0 mL of the stock solution and diluted it with enough water to make water to make 65.0 mL of a final solution.What is the concentration of KOH for the final solution?

What is the concentration of NO3- ions in a solution prepared by dissolving 30.0 g of Ca(NO3) 2 in enough water to produce 300.mL of solution?

What is the concentration (M) of a NaCl solution prepared by dissolving 5.3 g of NaCl in sufficient water to give 115 mL of solution?

What is the concentration (M) of CH3OH in a solution prepared by dissolving 12.5 g of CH3OH in sufficient water to give exactly 230 mL of solution?

What is the concentration of an AlCl3 solution if 150.mL of the solution contains 250.mg of Cl- ion?

How many grams of NaOH (MW = 40.0) are in 150.0 mL of a 0.300 M NaOH solution?

How many milliliters of a 9.0 M H2SO4 solution are needed to make 0.25 L of a 3.5 M solution?

How many milliliters of a stock solution of 13.5 M HNO3 would be needed to prepare 0.500 L of 0.500 M HNO3?

Pure acetic acid (HC2H3O2) is a liquid and is known as glacial acetic acid.Calculate the molarity of a solution prepared by dissolving 1.00 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.

If the reaction of phosphate ion with water is ignored,what is the total concentration of ions in a solution prepared by dissolving 4.00 g of K₃PO₄ in enough water to make 350.mL of solution?

How many grams of C OH must be added to water to prepare 325 mL of a solution that is 4.50 M CH₃OH?

How many grams of H₃PO₄ are in 255 mL of a 4.50 M solution of H₃PO₄?

How many aluminum ions are present in 65.5 mL of 0.210 M AlBr₃ solution?

A stock solution of HNO₃ is prepared and found to contain 14.9 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 iodide ions are present in 65.5 mL of 0.210 M AlI₃ solution?

How many ions are present in 30.0 mL of 0.600 M H₂CO₃ solution?

An FeCl₃ solution is 0.175 M.How many mL of a 0.175 M FeCl₃solution are needed to make 550.mL of a solution that is 0.300 M in Cl^- ion?

What is the concentration (M) of sodium ions in 4.57 L of a 2.35 M Li₃P solution?

What is the concentration of NO₃^- ions in a solution prepared by dissolving 30.0 g of Ca(NO₃) ₂ in enough water to produce 300.mL of solution?

What is the concentration (M) of CH₃OH in a solution prepared by dissolving 12.5 g of CH₃OH in sufficient water to give exactly 230 mL of solution?

What is the concentration of an AlCl₃ solution if 150.mL of the solution contains 250.mg of Cl^- ion?

How many milliliters of a 9.0 M H₂SO₄ solution are needed to make 0.25 L of a 3.5 M solution?

How many milliliters of a stock solution of 13.5 M HNO₃ would be needed to prepare 0.500 L of 0.500 M HNO₃?

Pure acetic acid (HC₂H₃O₂) is a liquid and is known as glacial acetic acid.Calculate the molarity of a solution prepared by dissolving 1.00 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.