Question 1

What is the concentration of sodium chloride in the final solution if   of   BaC  completely reacts and the total volume of the reaction is  , given the reaction:

A)0.226
B)0.0259
C)0.0519
D)0.667
E)none of the above

Accepted Answer

Step 1: Identify the limiting reactant: BaCl2

Step 2: Use stoichiometry to determine the number of moles of NaCl produced:

1 mole of BaCl2 produces 2 moles of NaCl

Therefore, 0.215 mole of BaCl2 produces 2*0.215 = 0.43 moles of NaCl

Step 3: Calculate the concentration of NaCl:

C = n/V

C = 0.43 moles / 8.28 L

C = 0.0519 M

Therefore, the concentration of NaCl in the final solution is 0.0519 M.

The answer is C.

Question 2

Which one of the following aqueous solutions would have the higher concentration of K⁺ (aq)ions? (Assume total solubility in water.) A)1.0 M KC₂H₃O₂ B)1.0 M KNO₃ C)1.0 M K₂CO₃ D)1.0 M K₃PO₄ E)All of these solutions have the same concentration of K⁺(aq).

Accepted Answer

The compound that leads to the formation of K⁺(aq) ions is K2S1U1, and D is the only option that contains K2S1U1. Therefore, it has the higher concentration of K⁺(aq) ions.

Question 3

How many milliliters of 0.755 M H₂SO₄ solution is needed to react with 55.0 mL of 2.50 M KOH solution? Given: 2 KOH (aq)+ H₂SO₄ (aq)→ 2 H₂O (l)+ K₂SO₄ (aq) A)51.9 mL B)182 mL C)91.1 mL D)17200 mL E)none of the above

Accepted Answer

To solve this problem, we need to use stoichiometry and the equation given to us. First, let's balance the equation: 2 KOH (aq) + H₂SO₄ (aq) → 2 H₂O (l) + K₂SO₄ (aq) We can see that the ratio of KOH and H₂SO₄ is 2:1, which means that for every 2 moles of KOH, we need 1 mole of H₂SO₄ to react. We can use the equation: moles = concentration (M) x volume (L) to find the number of moles of KOH we have: moles of KOH = concentration (2.50 M) x volume (55.0 mL / 1000 mL/L) = 0.1375 mol Since the ratio of KOH to H₂SO₄ is 2:1, we need half the number of moles of H₂SO₄: moles of H₂SO₄ = 0.1375 mol / 2 = 0.06875 mol Finally, we can use the equation: volume (L) = moles / concentration (M) to find the volume of 0.755 M H₂SO₄ we need: volume of H₂SO₄ = 0.06875 mol / 0.755 M = 0.0911 L = 91.1 mL Therefore, the best choice is C.

Question 4

Which solution below contains the highest total quantity of dissolved sodium ions? A)100.mL of 4.0 M NaCl B)75.0 mL of 3.0 M Na₂SO₄ C)50.0 mL of 8.0 M NaOH D)50.0 mL of 2.0 M Na₃PO₄ E)none of the above

Accepted Answer

The answer of Which solution below contains the highest total...

Question 5

In comparing a 0.25 molality aqueous NaCl solution to a 0.25 molality aqueous CaCl₂ solution: A)the NaCl solution has the higher boiling point and the lower freezing point. B)the CaCl₂ solution has the higher boiling point and the lower freezing point. C)the NaCl solution has the higher boiling point and the CaCl₂ solution has the lower freezing point. D)the CaCl₂ solution has the higher boiling point and the NaCl solution has the lower freezing point. E)both solutions have the same boiling point and the same freezing point.

Accepted Answer

The CaCl2 solution has a higher boiling point and a lower freezing point due to the van't Hoff factor. CaCl2 dissociates into three ions (Ca^2+ and 2 Cl^-), while NaCl dissociates into two ions (Na^+ and Cl^-), leading to a greater colligative effect for CaCl2.

Question 6

What volume of 9.00 M nitric acid is needed to make 6.50L of 1.25M solution?
A)903 mL
B)1.73 L
C)1.10 L
D)748 mL
E)none of the above

Accepted Answer

M1V1 = M2V29.00 M * V1 = 1.25 M * 6.50 LV1 = 0.903 L = 903 mL

Question 7

After you have completed the task of diluting a solution,which statement below must be TRUE?
A)The new solution has more volume but has a lower concentration than before.
B)The new solution has more volume but has a higher concentration than before.
C)The new solution has less volume but has a lower concentration than before.
D)The new solution has less volume but has a higher concentration than before.
E)none of the above

Accepted Answer

Dilution involves adding solvent to a solution to decrease its concentration. This means that the new solution will have more volume (due to the addition of solvent) but a lower concentration (due to the dilution). Therefore, choice A is the correct answer.

Question 8

What mass (in grams)of fructose (C₆H₁₂O₆)is contained in 125 mL of a 0.500 M fructose solution? A)11.3 B)11300 C)721 D)0.0625 E)180.

Accepted Answer

To find the mass of fructose in grams, first calculate the molar mass of fructose (C6H12O6). The molar mass is approximately 180.16 g/mol. Then, use the concentration and volume to find moles of fructose: 0.500 moles/L * 0.125 L = 0.0625 moles. Finally, multiply the moles by the molar mass to get the mass in grams: 0.0625 moles * 180.16 g/mol = 11.26 g, which rounds to 11.3 g.

Question 9

What are the ion concentrations in a 0.12 M solution of AlCl₃? A)0.36 M Al³⁺ ions and 0.12 M Cl^-ions B)0.12 M Al³⁺ ions and 0.040 M Cl^-ions C)0.12 M Al³⁺ ions and 0.36 M Cl^-ions D)0.040 M Al³⁺ ions and 0.040 M Cl^-ions E)none of the above

Accepted Answer

AlCl3 dissociates into one Al^3+ ion and three Cl^- ions for every formula unit that dissolves. Therefore, a 0.12 M solution of AlCl3 will have 0.12 M Al^3+ ions and 0.36 M Cl^- ions (0.12 M * 3).

Question 10

How much water would you ADD to 175 mL of 6.00 M hydrochloric acid to prepare a 2.00 M solution of this acid?
A)525 mL
B)350 mL
C)58.3 mL
D)700.mL
E)none of the above

Accepted Answer

To dilute the solution, use the formula M1V1 = M2V2. Here, M1 = 6.00 M, V1 = 175 mL, M2 = 2.00 M. Solving for V2 gives V2 = 525 mL. The amount of water to add is V2 - V1 = 525 mL - 175 mL = 350 mL.

Question 11

What molarity should the stock solution be if you want to dilute 25.0mL to 2.00L and have the final concentration be 0.103M ?
A)0.206 M
B)8.24 M
C)4.12 M
D)0.243 M
E)none of the above

Accepted Answer

First, use the dilution equation: 
M1V1 = M2V2 
where M1 is the initial concentration (unknown), V1 is the initial volume (25.0 mL = 0.025 L), M2 is the final concentration (0.103 M), and V2 is the final volume (2.00 L). 
Rearranging the equation to solve for M1: 
M1 = (M2V2)/V1 
M1 = (0.103 M)(2.00 L)/(0.025 L) 
M1 = 8.24 M 
Therefore, the stock solution should be 8.24 M in order to dilute 25.0 mL to 2.00 L and have the final concentration be 0.103 M.

Question 12

If you add 4.00 mL of pure water to 6.00 mL of 0.750 M NaCl solution,what is the concentration of sodium chloride in the diluted solution?
A)0.250 M
B)0.450 M
C)0.500 M
D)1.13 M
E)none of the above

Accepted Answer

Before dilution:
moles of NaCl = (6.00 mL) x (0.750 mol/L) = 4.5 mmol
After dilution:
total volume = 4.00 mL + 6.00 mL = 10.00 mL
moles of NaCl = 4.5 mmol
concentration = moles / volume = 4.5 mmol / 10.00 mL = 0.450 M

Question 13

How many grams of barium sulfate are produced if 25.34mL of 0.113M BaCl₂ completely react given the reaction: BaCl₂(aq)+ Na₂SO₄ (aq)→ BaSO₄ (s)+ 2NaCl (aq) A)5.90 B)26.3 C)1039 D)0.668 E)none of the above

Accepted Answer

The answer of How many grams of barium sulfate are...

Question 14

How many mL of 0.218 M sodium sulfate react with exactly   of   BaC  given the reaction:

A)13.1
B)5.52
C)24.6
D)2.86
E)none of the above

Accepted Answer

The answer of How many mL of 0.218 M sodium...

Question 15

A 0.15 M solution of BaCl₂ contains: A)0.15 M Ba²⁺ ions and 0.15 M Cl^- ions. B)0.15 M Ba²⁺ ions and 0.30 M Cl^- ions. C)0.30 M Ba²⁺ ions and 0.15 M Cl^- ions. D)0.30 M Ba²⁺ ions and 0.30 M Cl^- ions. E)none of the above

Accepted Answer

BaCl2 dissociates into one Ba2+ ion and two Cl- ions. Ba2+ reacts with S2- ions to form BaS solid, leaving half the Cl- ions remaining in solution. Therefore, the solution contains 0.15 M Ba2+ ions and 0.30 M Cl- ions. BaS dissociates into one Ba2+ ion and one S2- ion, so the solution also contains 0.15 M S2- ions. Therefore, choice B is correct, as it contains 0.15 M BaS and 0.30 M Cl-.

Question 16

How many liters of a 2.18 M solution can be made from   S?
A)0.832
B)1.81
C)0.252
D)1.20
E)none of the above

Accepted Answer

The answer of How many liters of a 2.18 M...

Question 17

What is the final concentration of a solution prepared by diluting 35.0 mL of 12.0 M HCl to a final volume of 1.20L ?
A)0.504 M
B)3.50 M
C)0.420 M
D)0.350 M
E)none of the above

Accepted Answer

The answer of What is the final concentration of a...

Question 18

Oftentimes solubility of a compound limits the concentration of the solution that can be prepared.Use the solubility data given with each compound shown below to determine which compound would allow the preparation of a 10.0 Molar solution. A)AgNO₃ (solubility = 122 g/100 g H₂O) B)KCl (solubility = 34.0 g/100 g H₂O) C)NaNO₃ (solubility = 89.0 g/100 g H₂O) D)NH₄Cl (solubility = 41.1 g/100 g H₂O) E)none of the above

Accepted Answer

The answer of Oftentimes solubility of a compound limits the...

Question 19

A 90.0 g sample of NaOH is dissolved in water and the solution is diluted to give a final volume of 3.00 liters.The molarity of the final solution is ________.
A)0.500 M
B)0.750 M
C)1.00 M
D)2.25 M
E)none of the above

Accepted Answer

The answer of A 90.0 g sample of NaOH is...

Question 20

What volume of 12.0 M HCl is required to make 75.0 mL of 3.50 M HCl?
A)21.9 mL
B)0.560 mL
C)257 mL
D)560.mL
E)none of the above

Accepted Answer

Use the dilution formula:

M1V1 = M2V2

Where:

M1 = 12.0 M (initial concentration)
V1 = unknown (volume of 12.0 M HCl to be used)
M2 = 3.50 M (final concentration)
V2 = 75.0 mL (final volume)

Solving for V1:

V1 = (M2 x V2) / M1

V1 = (3.50 M x 75.0 mL) / 12.0 M

V1 = 21.9 mL

Therefore, the answer is A) 21.9 mL

Quiz 13: Solutions

What is the concentration of sodium chloride in the final solution if of BaC completely reacts and the total volume of the reaction is , given the reaction:

Which one of the following aqueous solutions would have the higher concentration of K⁺ (aq) ions? (Assume total solubility in water.)

How many milliliters of 0.755 M H₂SO₄ solution is needed to react with 55.0 mL of 2.50 M KOH solution? Given: 2 KOH (aq) + H₂SO₄ (aq) → 2 H₂O (l) + K₂SO₄ (aq)

Which solution below contains the highest total quantity of dissolved sodium ions?

In comparing a 0.25 molality aqueous NaCl solution to a 0.25 molality aqueous CaCl₂ solution:

What volume of 9.00 M nitric acid is needed to make 6.50L of 1.25M solution?

After you have completed the task of diluting a solution,which statement below must be TRUE?

What mass (in grams) of fructose (C₆H₁₂O₆) is contained in 125 mL of a 0.500 M fructose solution?

What are the ion concentrations in a 0.12 M solution of AlCl₃?

How much water would you ADD to 175 mL of 6.00 M hydrochloric acid to prepare a 2.00 M solution of this acid?

What molarity should the stock solution be if you want to dilute 25.0mL to 2.00L and have the final concentration be 0.103M ?

If you add 4.00 mL of pure water to 6.00 mL of 0.750 M NaCl solution,what is the concentration of sodium chloride in the diluted solution?

How many grams of barium sulfate are produced if 25.34mL of 0.113M BaCl₂ completely react given the reaction: BaCl₂(aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2NaCl (aq)

How many mL of 0.218 M sodium sulfate react with exactly of BaC given the reaction:

A 0.15 M solution of BaCl₂ contains:

How many liters of a 2.18 M solution can be made from S?

What is the final concentration of a solution prepared by diluting 35.0 mL of 12.0 M HCl to a final volume of 1.20L ?

Oftentimes solubility of a compound limits the concentration of the solution that can be prepared.Use the solubility data given with each compound shown below to determine which compound would allow the preparation of a 10.0 Molar solution.

A 90.0 g sample of NaOH is dissolved in water and the solution is diluted to give a final volume of 3.00 liters.The molarity of the final solution is ________.

What volume of 12.0 M HCl is required to make 75.0 mL of 3.50 M HCl?