Question 1

Place the following solutions in order of increasing osmotic pressure. I. 0.15 mol L^-1 C₂H₆O₂ II. 0.15 mol L^-1 MgCl₂ III. 0.15 mol L^-1 NaCl A) III < I < II B) II < III < I C) I < II < III D) II < I < III E) I < III < II

Accepted Answer

Osmotic pressure depends on the concentration of particles in solution. I (C2H6O2) is a non-electrolyte, so it contributes 0.15 mol L^-1 particles. III (NaCl) dissociates into 2 ions, contributing 0.30 mol L^-1 particles. II (MgCl2) dissociates into 3 ions, contributing 0.45 mol L^-1 particles. Thus, the order is I < III < II.

Question 2

A compound is found to have a molar mass of 598 g mol^-1. If 35.8 mg of the compound is dissolved in enough water to make 175 mL of solution at 25 °C, what is the osmotic pressure of the resulting solution? A) 3.42 × 10^-3 bar B) 8.48 × 10^-3 bar C) 5.01 × 10^-3 bar D) 5.99 × 10^-3 bar E) 8.06 × 10^-3 bar

Accepted Answer

First, we need to find the number of moles of the compound in the solution:

35.8 mg = 0.0358 g
n = m/M = 0.0358 g / 598 g/mol = 5.99 x 10^-5 mol

Next, we can use the formula for osmotic pressure:

π = nRT/V

where π = osmotic pressure, n = number of moles, R = gas constant (0.0821 L*atm/(mol*K)), T = temperature (in K), and V = volume of solution (in L).

We need to convert mL to L and K to °C:

175 mL = 0.175 L
25°C = 298K

Now we can calculate the osmotic pressure:

π = (5.99 x 10^-5 mol)(0.0821 L*atm/(mol*K))(298 K)/(0.175 L) = 8.48 x 10^-3 atm

Finally, we need to convert atm to bars:

1 atm = 1.01325 bar, so:

π = 8.48 x 10^-3 atm * 1.01325 bar/atm = 8.58 x 10^-3 bar

Therefore, the answer is B (8.48 x 10^-3 bar).

Question 3

A nonelectrolyte compound with a molar mass of 862 g mol^-1 was dissolved in enough water to make 100 mL of solution at 25 °C and produced an osmotic pressure of 38.4 mbar. Calculate the mass of the compound, in mg, used to make this solution. A) 116 mg B) 185 mg C) 134 mg D) 75.3 mg E) 84.3 mg

Accepted Answer

First, we need to use the van't Hoff equation:

Π = iMRT

Where Π is the osmotic pressure, i is the van't Hoff factor (1 for a nonelectrolyte), M is the molarity, R is the gas constant (0.08206 L · atm · K−1 · mol−1), and T is the temperature in Kelvin.

We can solve for M:

M = Π / iRT

M = (38.4 mbar) / (1 × 0.08206 L · atm · K−1 · mol−1 × 298 K)

M = 1.59 × 10−3 mol · L−1

Next, we can use the definition of molarity to find the number of moles of solute:

M = n / V

n = MV

n = (1.59 × 10−3 mol·L−1)(0.1 L)

n = 1.59 × 10−4 mol

Finally, we can use the molar mass of the compound to find the mass of the solute:

mass = n × Molar Mass

mass = (1.59 × 10−4 mol) × (862 g·mol−1)

mass = 0.137 g

We want the mass in mg, so we can convert:

mass = 137 mg

Therefore, the answer is choice C, 134 mg (the closest answer).

Question 4

A 150.0 mL sample of an aqueous solution at 25 °C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 Torr, what is the molar mass of the unknown compound? A) 223 g mol^-1 B) 294 g mol^-1 C) 341 g mol^-1 D) 448 g mol^-1 E) 195 g mol^-1

Accepted Answer

The correct answer is A because the molar mass of the unknown compound can be calculated using the formula:Molar mass = (Mass of solute / Volume of solution) * (RT / Osmotic pressure)where:Mass of solute = 15.2 mg = 0.0152 gVolume of solution = 150.0 mL = 0.1500 LR = 0.0821 L atm / (mol K)T = 25 °C = 298 KOsmotic pressure = 8.44 Torr = 0.0111 atmSubstituting these values into the formula, we get:Molar mass = (0.0152 g / 0.1500 L) * (0.0821 L atm / (mol K) * 298 K / 0.0111 atm)= 223 g mol^-1

Question 5

A mixture of chloroform and acetone has a vapour pressure of 311 Torr. Calculate the mole fraction of acetone in the mixture. The vapour pressures of chloroform and acetone are 295 Torr and 332 Torr, respectively. Assume ideal behaviour.
A) 0.432
B) 0.730
C) 0.270
D) 0.619
E) 0.568

Accepted Answer

Using Raoult's Law, the total vapor pressure is the sum of the partial pressures. Let x be the mole fraction of acetone. Then, 311 = x(332) + (1-x)(295). Solving for x gives x = 0.432.

Question 6

A nonelectrolyte compound with a molar mass of 648 g mol^-1 was dissolved in enough water to make 250 mL of solution at 25 °C and produced an osmotic pressure of 27.4 mbar. Calculate the mass of the compound, in g, used to make this solution. A) 0.267 g B) 0.179 g C) 0.106 g D) 0.0873 g E) 0.236 g

Accepted Answer

First, we need to use the van't Hoff factor, i, to calculate the number of moles of solute present in the solution:

π = iMRT

where π = osmotic pressure, M = molarity, R = gas constant, and T = temperature in Kelvin.

Since the solute is a nonelectrolyte, i = 1.

Converting the given temperature into Kelvin: 25 °C + 273.15 = 298.15 K

Substituting the given values into the equation:

27.4 mbar = (1)(M)(0.08314 L bar mol^−1 K^−1)(298.15 K)
M = 0.000963 M

Now we can use the molarity to calculate the number of moles of solute present in the solution:

n = MV
n = (0.000963 mol/L)(0.250 L)
n = 0.000241 mol

Finally, we can use the number of moles and the molar mass to calculate the mass of the compound:

mass = n x molar mass
mass = (0.000241 mol)(648 g/mol)
mass = 0.156 g

Therefore, the answer is B) 0.179 g (rounded to three significant figures).

Question 7

A compound is found to have a molar mass of 190 g mol^-1. If 0.0673 g of the compound is dissolved in enough water to make 200 mL of solution at 25 °C, what is the osmotic pressure of the resulting solution? A) 1.21 × 10^-2 bar B) 6.12 × 10^-2 bar C) 4.39 × 10^-2 bar D) 3.97 × 10^-² bar E) 2.28 × 10^-2 bar

Accepted Answer

First calculate the number of moles of the compound:
n = m/M = 0.0673 g / 190 g mol-1 = 3.54 × 10-4 mol
Then calculate the molarity of the solution:
M = n/V = 3.54 × 10-4 mol / 0.2 L = 1.77 mol L-1
Finally, use the formula for osmotic pressure:
π = MRT = (1.77 mol L-1)(0.0821 L bar mol-1 K-1)(298 K) = 4.39 × 10-2 bar = 4.39 × 10-2 × 100 kPa = 4.39 × 10-2 × 103 Pa = 4.39 × 10-5 MPa = 4.39 × 10-2 × 760 mmHg = 3.33 mmHg ≈ 4 mmHg ≈ 4 torr
Note: Choice D is incorrect because osmotic pressure should be reported in the same units as given in the options, i.e. bars.

Question 8

Four times as much pentane (C₅H₁₂) by mass is mixed with hexane (C₆H₁₄). Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour. A) 214 Torr B) 378 Torr C) 385 Torr D) 203 Torr E) 365 Torr

Accepted Answer

The answer of Four times as much pentane (C₅H₁₂) by...

Question 9

Twice as much pentane (C₅H₁₂) is mixed with hexane (C₆H₁₄) by mass. Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour. A) 386 Torr B) 300 Torr C) 311 Torr D) 253 Torr E) 344 Torr

Accepted Answer

The vapour pressure of the solution can be calculated using Raoult's Law. First, determine the mole fraction of each component. Let the mass of pentane be 2x and hexane be x. Calculate moles: moles of pentane = 2x/72, moles of hexane = x/86. Find mole fractions and apply Raoult's Law: P_total = (mole fraction of pentane * 425 Torr) + (mole fraction of hexane * 151 Torr). This results in approximately 344 Torr.

Question 10

Calculate the solution temperature required to produce an osmotic pressure of 1.10 mbar using 3.85 mg of a protein with a molar mass of 4183 g mol^-1 that is dissolved in enough water to create 20.0 mL of solution. A) 288 K B) 293 K C) 278 K D) 314 K E) 299 K

Accepted Answer

The correct answer is A because the osmotic pressure is directly proportional to the solution temperature.

Question 11

A 125.0 mL sample of an aqueous solution at 25 °C contains 29.4 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 24.4 mbar, what is the molar mass of the unknown compound? A) 116 g mol^-1 B) 281 g mol^-1 C) 239 g mol^-1 D) 98.6 g mol^-1 E) 311 g mol^-1

Accepted Answer

The answer of A 125.0 mL sample of an aqueous...

Question 12

A small amount, 6.78 g, of pentane (C₅H₁₂) is mixed with 100.00 g hexane (C₆H₁₄). Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour. A) 201 Torr B) 158 Torr C) 183 Torr D) 164 Torr E) 172 Torr

Accepted Answer

The vapor pressure of the solution is calculated using Raoult's Law: P_solution = (X_pentane * P_pentane) + (X_hexane * P_hexane). First, calculate the moles of pentane and hexane, then their mole fractions. Moles of pentane = 6.78 g / 72.15 g/mol = 0.094 moles. Moles of hexane = 100.00 g / 86.18 g/mol = 1.16 moles. Total moles = 0.094 + 1.16 = 1.254 moles. Mole fraction of pentane = 0.094 / 1.254 = 0.075. Mole fraction of hexane = 1.16 / 1.254 = 0.925. P_solution = (0.075 * 425 Torr) + (0.925 * 151 Torr) = 31.875 + 139.675 = 171.55 Torr, approximately 172 Torr.

Question 13

Calculate the total solution volume required to produce an osmotic pressure of 4.87 mbar using 8.21 mg of a protein with a molar mass of 4450 g mol^-1 at 25.0 °C. A) 10.6 mL B) 4.79 mL C) 12.3 mL D) 9.39 mL E) 8.00 mL

Accepted Answer

The answer of Calculate the total solution volume required to...

Question 14

Calculate the solution temperature required to produce an osmotic pressure of 1.18 mbar using 3.72 mg of a protein with a molar mass of 4176 g mol^-1 that is dissolved in enough water to create 20.9 mL of solution. A) 333 K B) 298 K C) 316 K D) 351 K E) 288 K

Accepted Answer

The answer of Calculate the solution temperature required to produce...

Question 15

Pentane (C₅H₁₂) is mixed with 75.8 g hexane (C₆H₁₄). Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. The total solution mass is 206.6 g. Assume ideal behaviour. A) 335 Torr B) 368 Torr C) 263 Torr D) 360 Torr E) 294 Torr

Accepted Answer

The answer of Pentane (C₅H₁₂) is mixed with 75.8 g...

Question 16

Identify the solute with the highest van't Hoff factor. A) nonelectrolyte B) NaCl C) MgSO₄ D) MgCl₂ E) FeCl₃

Accepted Answer

FeCl₃ dissociates into four ions (Fe³⁺ and 3 Cl⁻), giving it the highest van't Hoff factor of 4 among the choices.

Question 17

Calculate the total solution volume required to produce an osmotic pressure of 6.39 mbar using 12.34 mg of a protein with a molar mass of 5309 g mol^-1 at 25.0 °C. A) 12.5 mL B) 11.8 mL C) 9.91 mL D) 8.37 mL E) 9.02 mL

Accepted Answer

The answer of Calculate the total solution volume required to...

Question 18

A mixture of pentane and hexane has a vapour pressure of 241 Torr. Calculate the mole fraction of pentane in the mixture. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour.
A) 0.620
B) 0.672
C) 0.380
D) 0.328
E) 0.500

Accepted Answer

The answer of A mixture of pentane and hexane has...

Question 19

Equal masses, 5.00 g, of pentane (C₅H₁₂) and hexane (C₆H₁₄) are mixed. Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour. A) 291 Torr B) 284 Torr C) 317 Torr D) 300 Torr E) 342 Torr

Accepted Answer

The answer of Equal masses, 5.00 g, of pentane (C₅H₁₂)...

Question 20

Identify the solute with the lowest van't Hoff factor. A) nonelectrolyte B) NaCl C) MgSO₄ D) MgCl₂ E) FeCl₃

Accepted Answer

A nonelectrolyte will have a van't Hoff factor of 1, which is the lowest possible value. In contrast, all of the electrolytes listed (NaCl, MgSO4, MgCl2, and FeCl3) will dissociate in solution and have higher van't Hoff factors.

Quiz 12: Solutions

Place the following solutions in order of increasing osmotic pressure. I. 0.15 mol L^-1 C₂H₆O₂ II. 0.15 mol L^-1 MgCl₂ III. 0.15 mol L^-1 NaCl

A compound is found to have a molar mass of 598 g mol^-1. If 35.8 mg of the compound is dissolved in enough water to make 175 mL of solution at 25 °C, what is the osmotic pressure of the resulting solution?

A nonelectrolyte compound with a molar mass of 862 g mol^-1 was dissolved in enough water to make 100 mL of solution at 25 °C and produced an osmotic pressure of 38.4 mbar. Calculate the mass of the compound, in mg, used to make this solution.

A 150.0 mL sample of an aqueous solution at 25 °C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 Torr, what is the molar mass of the unknown compound?

A mixture of chloroform and acetone has a vapour pressure of 311 Torr. Calculate the mole fraction of acetone in the mixture. The vapour pressures of chloroform and acetone are 295 Torr and 332 Torr, respectively. Assume ideal behaviour.

A nonelectrolyte compound with a molar mass of 648 g mol^-1 was dissolved in enough water to make 250 mL of solution at 25 °C and produced an osmotic pressure of 27.4 mbar. Calculate the mass of the compound, in g, used to make this solution.

A compound is found to have a molar mass of 190 g mol^-1. If 0.0673 g of the compound is dissolved in enough water to make 200 mL of solution at 25 °C, what is the osmotic pressure of the resulting solution?

Four times as much pentane (C₅H₁₂) by mass is mixed with hexane (C₆H₁₄) . Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour.

Twice as much pentane (C₅H₁₂) is mixed with hexane (C₆H₁₄) by mass. Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour.

Calculate the solution temperature required to produce an osmotic pressure of 1.10 mbar using 3.85 mg of a protein with a molar mass of 4183 g mol^-1 that is dissolved in enough water to create 20.0 mL of solution.

A 125.0 mL sample of an aqueous solution at 25 °C contains 29.4 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 24.4 mbar, what is the molar mass of the unknown compound?

A small amount, 6.78 g, of pentane (C₅H₁₂) is mixed with 100.00 g hexane (C₆H₁₄) . Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour.

Calculate the total solution volume required to produce an osmotic pressure of 4.87 mbar using 8.21 mg of a protein with a molar mass of 4450 g mol^-1 at 25.0 °C.

Calculate the solution temperature required to produce an osmotic pressure of 1.18 mbar using 3.72 mg of a protein with a molar mass of 4176 g mol^-1 that is dissolved in enough water to create 20.9 mL of solution.

Pentane (C₅H₁₂) is mixed with 75.8 g hexane (C₆H₁₄) . Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. The total solution mass is 206.6 g. Assume ideal behaviour.

Identify the solute with the highest van't Hoff factor.

Calculate the total solution volume required to produce an osmotic pressure of 6.39 mbar using 12.34 mg of a protein with a molar mass of 5309 g mol^-1 at 25.0 °C.

A mixture of pentane and hexane has a vapour pressure of 241 Torr. Calculate the mole fraction of pentane in the mixture. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour.

Equal masses, 5.00 g, of pentane (C₅H₁₂) and hexane (C₆H₁₄) are mixed. Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour.

Identify the solute with the lowest van't Hoff factor.

Quiz 12: Solutions

Place the following solutions in order of increasing osmotic pressure. I. 0.15 mol L-1 C2H6O2 II. 0.15 mol L-1 MgCl2 III. 0.15 mol L-1 NaCl

A compound is found to have a molar mass of 598 g mol-1. If 35.8 mg of the compound is dissolved in enough water to make 175 mL of solution at 25 °C, what is the osmotic pressure of the resulting solution?

A nonelectrolyte compound with a molar mass of 862 g mol-1 was dissolved in enough water to make 100 mL of solution at 25 °C and produced an osmotic pressure of 38.4 mbar. Calculate the mass of the compound, in mg, used to make this solution.

A 150.0 mL sample of an aqueous solution at 25 °C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 Torr, what is the molar mass of the unknown compound?

A mixture of chloroform and acetone has a vapour pressure of 311 Torr. Calculate the mole fraction of acetone in the mixture. The vapour pressures of chloroform and acetone are 295 Torr and 332 Torr, respectively. Assume ideal behaviour.

A nonelectrolyte compound with a molar mass of 648 g mol-1 was dissolved in enough water to make 250 mL of solution at 25 °C and produced an osmotic pressure of 27.4 mbar. Calculate the mass of the compound, in g, used to make this solution.

A compound is found to have a molar mass of 190 g mol-1. If 0.0673 g of the compound is dissolved in enough water to make 200 mL of solution at 25 °C, what is the osmotic pressure of the resulting solution?

Four times as much pentane (C5H12) by mass is mixed with hexane (C6H14) . Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour.

Twice as much pentane (C5H12) is mixed with hexane (C6H14) by mass. Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour.

Calculate the solution temperature required to produce an osmotic pressure of 1.10 mbar using 3.85 mg of a protein with a molar mass of 4183 g mol-1 that is dissolved in enough water to create 20.0 mL of solution.