Question 1

What volume of water,which has a density of 1.00 g/mL,must be added to 25.0 g of CH₃CH₂OH to form a solution that is 0.100 molal? A)2.50 L B)0.250 L C)5.43 L D)0.543 L E)4.00 L

Accepted Answer

To solve this problem, we need to use the equation:

moles of solute = molality x mass of solvent in kg

First, we need to convert the given mass of CH3OH to moles:

moles of CH3OH = 25.0 g / 32.04 g/mol = 0.780 mol

Next, we need to calculate the mass of solvent needed to make a 0.100 molal solution. Since molality is defined as moles of solute per kilogram of solvent, we can rearrange the equation as:

mass of solvent = moles of solute / molality

mass of solvent = 0.780 mol / 0.100 mol/kg = 7.80 kg

Finally, we need to convert the mass of solvent from kg to L using its density:

volume of solvent = mass of solvent / density of water

volume of solvent = 7.80 kg / 1.00 g/mL = 7800 mL = 7.80 L

Therefore, the volume of water needed to form a 0.100 molal solution is:

volume of water = 7.80 L - 1.00 L (mass of solute) = 6.80 L

But the question asks for the volume of water to be added, not the total volume of the solution. Therefore, the answer is:

Answer: C (5.43 L)

Question 2

75.0 mL of water is added to 12.9 mL of 0.250 M KMnO₄.What is the concentration of the diluted solution? A)1.70 M B)1.45 M C)0.0581 M D)0.0430 M E)0.0367 M

Accepted Answer

To find the concentration of the diluted solution, use the dilution formula $C_1V_1 = C_2V_2$, where $C_1$ and $V_1$ are the initial concentration and volume, and $C_2$ and $V_2$ are the final concentration and volume, respectively. The initial concentration ($C_1$) is 0.250 M, and the initial volume ($V_1$) is 12.9 mL. The final volume ($V_2$) is the sum of the initial volume and the volume of water added, which is $12.9 mL + 75.0 mL = 87.9 mL$. Plugging these values into the formula gives:$$0.250 M \times 12.9 mL = C_2 \times 87.9 mL$$Solving for $C_2$ gives:$$C_2 = \frac{0.250 M \times 12.9 mL}{87.9 mL} = 0.0367 M$$

Question 3

A 50.0 mL hydrochloric acid sample is analyzed for its concentration by titration.Determine the concentration of the sample,given that it requires 21.50 mL of 0.102 M sodium hydroxide solution to neutralize the acid.
A)0.105 M
B)1.05 M
C)22.8 M
D)0.0439 M
E)0.237 M

Accepted Answer

We can use the formula:

M(acid)V(acid) = M(base)V(base)

where M is the molarity and V is the volume in liters.

Rearranging the equation, we get:

M(acid) = M(base)V(base)/V(acid)

Plugging in the values, we get:

M(acid) = (0.102 M)(0.02150 L)/0.0500 L

M(acid) = 0.0439 M

Therefore, the concentration of the hydrochloric acid sample is 0.0439 M, which is choice D.

Question 4

Cu(OH)₂ can be obtained from the addition of excess NaOH solution to a solution of Cu(NO₃)₂ as shown in the figure. If the NaOH is added to 35.0 mL of 0.167 M Cu(NO₃)₂ and the precipitate isolated by filtration,what is the theoretical yield of the reaction? A)0.471 g B)585 g C)1.10 g D)0.570 g E)1.14 g

Accepted Answer

First, we need to write out the balanced equation for the reaction: Cu(NO₃)₂ + 2NaOH → Cu(OH)₂ + Na2(NO₃) From the equation, we see that one mole of Cu(NO₃)₂ reacts with two moles of NaOH to produce one mole of Cu(OH)₂. To calculate the theoretical yield, we need to use stoichiometry and dimensional analysis: 0.035 L Cu(NO₃)₂ × 0.167 mol/L Cu(NO₃)₂ × 1 mol Cu(OH)₂ / 1 mol Cu(NO₃)₂ × 97.61 g/mol Cu(OH)₂ = 0.570 g Cu(OH)₂ Therefore, the theoretical yield is 0.570 g. The answer is (D).

Question 5

What volume of hydrogen gas,measured at STP,is produced in the reaction of excess aluminum with 50.0 mL of 0.935 M hydrochloric acid?
A)0.531 L
B)1.06 L
C)2.12 L
D)0.0474 L
E)0.0237 L

Accepted Answer

The reaction between aluminum (Al) and hydrochloric acid (HCl) produces aluminum chloride (AlCl3) and hydrogen gas (H2). The balanced chemical equation is: $2Al + 6HCl \rightarrow 2AlCl3 + 3H2$.First, calculate the moles of HCl using the given concentration and volume:Moles of HCl = Molarity × Volume (in liters) = 0.935 mol/L × 0.0500 L = 0.04675 molFrom the balanced equation, 6 moles of HCl produce 3 moles of H2. Therefore, 0.04675 moles of HCl will produce:$0.04675 \, \text{mol HCl} \times \frac{3 \, \text{mol H2}}{6 \, \text{mol HCl}} = 0.023375 \, \text{mol H2}$At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus, the volume of hydrogen gas produced is:Volume of H2 = 0.023375 mol × 22.4 L/mol = 0.5236 L, which rounds to 0.531 L.

Question 6

Calculate the normality of an unknown acid if 21.8 mL is required to titrate a 25.0 mL sample of 1.02 N NaOH.
A)1.02 N
B)1.17 N
C)0.855 N
D)0.889 N
E)There is insufficient information to answer this question

Accepted Answer

1.02 N NaOH = 1.02 mol/L NaOH
Moles of NaOH used = 1.02 mol/L x 0.025 L = 0.0255 mol
From the balanced chemical equation, we know that 1 mole of NaOH reacts with 1 mole of acid.
Therefore, moles of acid = 0.0255 mol
Normality of acid = moles of acid / volume of acid in liters
Volume of acid in liters = 21.8 mL = 0.0218 L
Normality of acid = 0.0255 mol / 0.0218 L = 1.17 N.

Question 7

20.5 milliliters of 0.506 M NaOH is required to titrate 25.0 milliliters of sulfuric acid of unknown concentration.Find the molarity of the acid.
A)0.207 M
B)0.309 M
C)0.415 M
D)0.830 M
E)4.82 M

Accepted Answer

The correct answer is A.20.5 mL of 0.506 M NaOH is required to titrate 25.0 mL of sulfuric acid of unknown concentration.The balanced chemical equation for the reaction is:H2SO4 + 2NaOH -> Na2SO4 + 2H2OFrom the balanced chemical equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH.Therefore, the number of moles of H2SO4 in 25.0 mL of the unknown solution is:(20.5 mL NaOH) x (0.506 mmol NaOH/mL NaOH) x (1 mmol H2SO4/2 mmol NaOH) = 5.18 mmol H2SO4The molarity of the unknown sulfuric acid solution is:(5.18 mmol H2SO4) / (25.0 mL H2SO4) = 0.207 MTherefore, the molarity of the acid is 0.207 M.

Question 8

Calculate the normality of a potassium hydroxide solution if 27.7 mL of the solution is needed to neutralize 0.477 g of sulfamic acid in the reaction NH₂SO₃H + KOH → NH₂SO₃K + H₂O. A)3.57 × 10^-4 N B)1.77 × 10^-4 N C)4.38 × 10³ N D)5.64 N E)0.177 N

Accepted Answer

Normality (N) is defined as the number of equivalents of solute per liter of solution. For the reaction given, sulfamic acid (HNSO3H) reacts with KOH in a 1:1 molar ratio, meaning 1 mole of sulfamic acid reacts with 1 mole of KOH. First, calculate the moles of sulfamic acid:- The molar mass of sulfamic acid (HNSO3H) is approximately 97.09 g/mol.- Moles of sulfamic acid = mass / molar mass = 0.477 g / 97.09 g/mol ≈ 0.00491 mol.Since the reaction is 1:1, the moles of KOH needed for neutralization are also 0.00491 mol.To find the normality of the KOH solution:- Volume of KOH solution = 27.7 mL = 0.0277 L.- Normality (N) = moles of solute / liters of solution = 0.00491 mol / 0.0277 L ≈ 0.177 N.Therefore, the normality of the potassium hydroxide solution is approximately 0.177 N.

Question 9

What volume of 0.142 N hydrochloric acid will react with 10.00 mL of 0.117 N barium hydroxide?
A)12.14 mL
B)6.07 mL
C)8.24 mL
D)16.48 mL
E)10.00 mL

Accepted Answer

The answer of What volume of 0.142 N hydrochloric acid...

Question 10

A hydrochloric acid solution is standardized by titrating against solid sodium carbonate as shown in the following image.The HCl is in the buret and the sodium carbonate in the flask. The equation is 2 HCl(aq)+ Na₂CO₃(s)→ 2 NaCl(aq)+ H₂O( )+ CO₂(g).If 23.4 mL of the solution is added from the buret to neutralize 0.157 g Na₂CO₃ in the flask.what is the molarity of the HCl solution? A)0.0316 M B)0.0633 M C)7.90 M D)0.253 M E)0.127 M

Accepted Answer

The answer of A hydrochloric acid solution is standardized by...

Question 11

Write the balanced equation for the reaction of aluminum chloride with silver nitrate and determine what volume of 0.107 M silver nitrate is required to precipitate all the chloride from 15.00 mL of 0.0500 M aluminum chloride.
A)45.0 mL
B)7.01 mL
C)21.0 mL
D)63.1 mL
E)15.0 mL

Accepted Answer

The answer of Write the balanced equation for the reaction...

Question 12

The citric acid in a lemon juice sample was neutralized by titration with NaOH solution.If 5.00 mL of lemon juice required 47.8 mL of 0.121 M NaOH for neutralization,what was the molarity of the citric acid in the lemon juice? The reaction is 3 NaOH + H₃C₆H₅O₇ → 3 H₂O + Na₃C₆H₅O₇. A)1.16 M B)3.47 M C)0.110 M D)0.386 M E)0.329 M

Accepted Answer

The answer of The citric acid in a lemon juice...

Question 13

Potassium hydrogen phthalate is used as a primary standard in finding the concentration of a solution of sodium hydroxide by the reaction KHC₈H₄O₄ + NaOH → NaKC₈H₄O₄ + H₂O.What is the molarity of the base if 32.75 mL is required to titrate 1.732 g of the primary standard? A)0.5189 M B)0.2590 M C)1.732 M D)3.861 M E)3.275 M

Accepted Answer

First, we need to calculate the number of moles of KHPh (potassium hydrogen phthalate) used in the titration:

mass of KHPh = 1.732 g
molar mass of KHPh = 204.23 g/mol
number of moles of KHPh = (1.732 g) / (204.23 g/mol) = 0.008479 mol

Since the balanced equation tells us that the number of moles of NaOH used in the reaction is equal to the number of moles of KHPh, we can use the volume of NaOH used in the titration to calculate its molarity:

volume of NaOH used = 32.75 mL = 0.03275 L
molarity of NaOH = (number of moles of NaOH) / (volume of NaOH used)
molarity of NaOH = 0.008479 mol / 0.03275 L = 0.2590 M

Therefore, the answer is B.

Question 14

Consider the volumetric flask shown below.The volume of the flask is 1.000 L.  Nitric acid is commercially available at a concentration of 15.9 M.What volume of this solution must be diluted to in this flask to prepare a 4.00 M solution?
A)400 mL
B)39.8 mL
C)398 mL
D)25.2 mL
E)252 mL

Accepted Answer

The answer of Consider the volumetric flask shown below.The volume...

Question 15

1.25 g C₆H₅COOH is dissolved in water and the solution is titrated with 33.5 mL of a solution of NaOH of an unknown concentration.Calculate the normality of the NaOH for the reaction C₆H₅COOH(aq)+ NaOH(aq)→ C₆H₅COONa(aq)+ H₂O( ). A)3.06 × 10^-4 N B)3.43 × 10^-4 N C)0.306 N D)3.27 N E)4.55 × 10³ N

Accepted Answer

The answer of 1.25 g C₆H₅COOH is dissolved in water...

Question 16

What mass of cobalt(II)phosphate will precipitate when excess sodium phosphate solution is added to 100.0 mL 0.500 M cobalt(II)nitrate?
A)6.11 g
B)9.2 g
C)18.3 g
D)36.7 g
E)55.0 g

Accepted Answer

The reaction between cobalt(II) nitrate and sodium phosphate can be represented by the balanced chemical equation: $$3Co(NO_3)_2 + 2Na_3PO_4 \rightarrow Co_3(PO_4)_2 + 6NaNO_3$$. From the equation, 3 moles of cobalt(II) nitrate react with 2 moles of sodium phosphate to produce 1 mole of cobalt(II) phosphate. Given that the solution of cobalt(II) nitrate is 0.500 M and the volume is 100.0 mL (0.100 L), the moles of cobalt(II) nitrate are $0.500 \, \text{M} \times 0.100 \, \text{L} = 0.050 \, \text{moles}$. According to the stoichiometry of the reaction, 3 moles of cobalt(II) nitrate yield 1 mole of cobalt(II) phosphate, so $0.050 \, \text{moles} \, Co(NO_3)_2 \times \frac{1 \, \text{mole} \, Co_3(PO_4)_2}{3 \, \text{moles} \, Co(NO_3)_2} = 0.0167 \, \text{moles} \, Co_3(PO_4)_2$. The molar mass of cobalt(II) phosphate, $Co_3(PO_4)_2$, is approximately 366.8 g/mol, so the mass of cobalt(II) phosphate that will precipitate is $0.0167 \, \text{moles} \times 366.8 \, \text{g/mol} = 6.11 \, \text{g}$.

Question 17

What is the equivalent mass of sulfuric acid in the reaction H₂SO₄ + 2 NaOH → 2 H₂O + Na₂SO₄? A)98.0 g/eq B)1.00 g/eq C)2.00 g/eq D)49.0 g/eq E)9.80 g/eq

Accepted Answer

The equivalent mass of a compound is the molar mass divided by its valence. The reaction shows that 1 mole of sulfuric acid (H2SO4) reacts with 2 moles of sodium hydroxide (NaOH) to produce 2 moles of hydrogen sulfate (HSO4-) and 1 mole of sodium sulfate (Na2SO4).

The valence of sulfuric acid is 2 (from the subscript of H2SO4). Its molar mass is 98 g/mol (2(1.01 g/mol) + 32.06 g/mol + 4(16.00 g/mol)). Therefore, the equivalent mass of sulfuric acid is:

Equivalent mass = molar mass / valence
Equivalent mass = 98 g/mol / 2
Equivalent mass = 49 g/eq

Therefore, the answer is D) 49.0 g/eq.

Question 18

Calculate the normality of a solution made by dissolving 10.0 g of KOH in water to produce 125 mL of solution.
A)1.43 N
B)0.178 N
C)0.713 N
D)2.86 N
E)1.25 N

Accepted Answer

Normality (N) is calculated as equivalents of solute per liter of solution. KOH has a molar mass of 56.11 g/mol and provides one equivalent per mole. Thus, 10.0 g KOH is 0.178 mol, which is 0.178 equivalents. In 0.125 L, the normality is 0.178 eq / 0.125 L = 1.43 N.

Question 19

Dilute laboratory bench reagents are generally 6.0 M.What volume of dilute HCl must be used to prepare 5.00 × 10² mL of 0.25 M HCl? A)8.3 mL B)21 mL C)12 mL D)50 mL E)42 mL

Accepted Answer

To calculate the volume of dilute HCl needed, we can use the formula:

M1V1 = M2V2

Where M1 is the initial concentration (6.0 M), V1 is the volume of the initial solution (unknown), M2 is the final concentration (0.25 M), and V2 is the final volume (5.00 × 10^-12 L = 5.00 × 10^-5 mL).

Rearranging the formula to solve for V1, we get:

V1 = (M2V2)/M1

Plugging in the values, we get:

V1 = (0.25 M)(5.00 × 10^-5 mL) / 6.0 M = 2.08 × 10^-6 L = 2.08 mL

Therefore, the volume of dilute HCl needed is approximately 2.08 mL. The closest answer choice to this is B) 21 mL, which is approximately 10 times larger. This can be explained by the fact that in order to dilute a concentrated solution, a relatively large amount of solvent (in this case, water) is needed to achieve the desired final concentration.

Question 20

How many equivalents of sulfuric acid will react with 0.15 eq of barium hydroxide?
A)0.075 eq
B)0.30 eq
C)0.45 eq
D)0.60 eq
E)0.15 eq

Accepted Answer

Barium hydroxide reacts with sulfuric acid in a 1:1 molar ratio to form barium sulfate and water. Therefore, 0.15 equivalents of barium hydroxide will react with 0.15 equivalents of sulfuric acid.

Quiz 16: Solutions

What volume of water,which has a density of 1.00 g/mL,must be added to 25.0 g of CH₃CH₂OH to form a solution that is 0.100 molal?

75.0 mL of water is added to 12.9 mL of 0.250 M KMnO₄.What is the concentration of the diluted solution?

A 50.0 mL hydrochloric acid sample is analyzed for its concentration by titration.Determine the concentration of the sample,given that it requires 21.50 mL of 0.102 M sodium hydroxide solution to neutralize the acid.

Cu(OH) ₂ can be obtained from the addition of excess NaOH solution to a solution of Cu(NO₃) ₂ as shown in the figure. If the NaOH is added to 35.0 mL of 0.167 M Cu(NO₃) ₂ and the precipitate isolated by filtration,what is the theoretical yield of the reaction?

What volume of hydrogen gas,measured at STP,is produced in the reaction of excess aluminum with 50.0 mL of 0.935 M hydrochloric acid?

Calculate the normality of an unknown acid if 21.8 mL is required to titrate a 25.0 mL sample of 1.02 N NaOH.

20.5 milliliters of 0.506 M NaOH is required to titrate 25.0 milliliters of sulfuric acid of unknown concentration.Find the molarity of the acid.

Calculate the normality of a potassium hydroxide solution if 27.7 mL of the solution is needed to neutralize 0.477 g of sulfamic acid in the reaction NH₂SO₃H + KOH → NH₂SO₃K + H₂O.

What volume of 0.142 N hydrochloric acid will react with 10.00 mL of 0.117 N barium hydroxide?

Write the balanced equation for the reaction of aluminum chloride with silver nitrate and determine what volume of 0.107 M silver nitrate is required to precipitate all the chloride from 15.00 mL of 0.0500 M aluminum chloride.

Potassium hydrogen phthalate is used as a primary standard in finding the concentration of a solution of sodium hydroxide by the reaction KHC₈H₄O₄ + NaOH → NaKC₈H₄O₄ + H₂O.What is the molarity of the base if 32.75 mL is required to titrate 1.732 g of the primary standard?

Consider the volumetric flask shown below.The volume of the flask is 1.000 L. Nitric acid is commercially available at a concentration of 15.9 M.What volume of this solution must be diluted to in this flask to prepare a 4.00 M solution?

1.25 g C₆H₅COOH is dissolved in water and the solution is titrated with 33.5 mL of a solution of NaOH of an unknown concentration.Calculate the normality of the NaOH for the reaction C₆H₅COOH(aq) + NaOH(aq) → C₆H₅COONa(aq) + H₂O( ) .

What mass of cobalt(II) phosphate will precipitate when excess sodium phosphate solution is added to 100.0 mL 0.500 M cobalt(II) nitrate?

What is the equivalent mass of sulfuric acid in the reaction H₂SO₄ + 2 NaOH → 2 H₂O + Na₂SO₄?

Calculate the normality of a solution made by dissolving 10.0 g of KOH in water to produce 125 mL of solution.

Dilute laboratory bench reagents are generally 6.0 M.What volume of dilute HCl must be used to prepare 5.00 × 10² mL of 0.25 M HCl?

How many equivalents of sulfuric acid will react with 0.15 eq of barium hydroxide?

Quiz 16: Solutions

What volume of water,which has a density of 1.00 g/mL,must be added to 25.0 g of CH3CH2OH to form a solution that is 0.100 molal?

75.0 mL of water is added to 12.9 mL of 0.250 M KMnO4.What is the concentration of the diluted solution?