Question 1

At 25°C,what is the osmotic pressure of a homogeneous solution consisting of 18.0 g urea (CON₂H₄)diluted with water to 3.00 L? (R = 0.0821 L.atm/mol.K)

Accepted Answer

First, we need to find the molarity of the solution:
n(urea) = mass/molecular weight
n(urea) = 18.0 g / 60.06 g/mol = 0.2996 mol

Molarity = n/V
Molarity = 0.2996 mol / 3.00 L = 0.0999 M

Now, we can use the van't Hoff factor (i) to calculate the effective concentration of particles in solution:

i = 1 (urea does not dissociate in water)

π = iMRT
π = (1)(0.0999 M)(0.0821 L.atm/mol.K)(298 K)
π = 2.44 atm

Therefore, the osmotic pressure of the solution is 2.44 atm, which corresponds to choice B.

Question 2

An aqueous solution is composed of 7.50 g NaCl (MM = 58.44 g/mol)diluted to 0.100 L.Calculate the osmotic pressure of the solution at 298 K.(R = 0.0821 L.atm/mol.K)

Accepted Answer

The answer of An aqueous solution is composed of 7.50...

Question 3

Equal masses of water and ethylene glycol (C₂H₆O₂)are mixed.At what temperature will the mixture freeze? The molal freezing point constant for water is −1.86°C/m.

Accepted Answer

The answer of Equal masses of water and ethylene glycol...

Question 4

The molal freezing point constant for water is -1.86°C/m.At what temperature will a solution containing 8.27 g CaCl₂ and 45.0 g H₂O begin to freeze? Assume that no ion-pairing occurs between Ca²⁺ and Cl^-.

Accepted Answer

The freezing point depression of a solution can be calculated using the formula ΔT_f = i * K_f * m, where ΔT_f is the change in freezing point, i is the van't Hoff factor (which represents the number of particles the solute dissociates into), K_f is the molal freezing point depression constant for the solvent (in this case, water, which is -1.86°C/m), and m is the molality of the solution.For CaCl2:- The molar mass of CaCl2 is approximately 40.08 (Ca) + 35.45*2 (Cl) = 110.98 g/mol.- The amount of CaCl2 in moles is 8.27 g / 110.98 g/mol = 0.0745 mol.- The van't Hoff factor (i) for CaCl2, which dissociates into 3 ions (Ca^2+ and 2 Cl^-), is 3.- The mass of water in kilograms is 45.0 g / 1000 = 0.045 kg.- The molality (m) of the solution is 0.0745 mol / 0.045 kg = 1.655 mol/kg.Using the formula:ΔT_f = i * K_f * m = 3 * (-1.86°C/m) * 1.655 mol/kg = -9.24°C.This means the solution will begin to freeze at 0°C - 9.24°C = -9.24°C.

Question 5

Which of the following electrolytes is likely to have a van't Hoff factor equal to 3?

Accepted Answer

CaI₂ dissociates into three ions: one Ca²⁺ and two I^-, resulting in a van't Hoff factor of 3.

Question 6

A 0.230 m solution of an unknown electrolyte depresses the freezing point of water by 0.821°C.What is the Van't Hoff factor for this electrolyte? The freezing point depression constant for water is 1.86°C/m.

Accepted Answer

The answer of A 0.230 m solution of an unknown...

Question 7

A solution is prepared by dissolving 4.21 g of a nonelectrolyte in 50.0 g of water.If the boiling point increases by 0.203°C,what is the molar mass of the solute? The boiling point elevation constant for water is 0.512°C/m.

Accepted Answer

First, we can calculate the molality (b) of the solution using the formula:

ΔT = Kb * b

where ΔT is the boiling point elevation, Kb is the boiling point elevation constant for water, and b is the molality of the solution.

Substituting the given values, we get:

0.203°C = 0.512°C/m * b

b = 0.3975 m

Next, we can calculate the moles of the solute using the formula:

moles solute = m * kg solvent / M

where m is the molality of the solution, kg solvent is the mass of the solvent (in kg), and M is the molar mass of the solute.

Substituting the given values, we get:

moles solute = 0.3975 * 0.05 / M

moles solute = 0.019875 / M

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

molar mass = mass solute / moles solute

molar mass = 4.21 g / 0.019875 mol

molar mass = 211.4 g/mol

Rounding off to the nearest whole number, we get:

molar mass = 212 g/mol

Therefore, the correct answer is D) 212 g/mol.

Question 8

What is the boiling point of a solution containing 0.80 g caffeine,C₈H₁₀N₄O₂,dissolved in 13.20 g benzene? The boiling point of pure benzene is 80.1°C and the molal boiling point constant,K_b,is 2.53°C/m.

Accepted Answer

First, we need to calculate the molality of the solution:
molality (m) = (0.80 g / 194.19 g/mol) / (13.20 g / 78.11 g/mol) = 0.0574 mol/kg

Next, we can use the formula for boiling point elevation:

ΔTb = Kb x m

where ΔTb is the boiling point elevation, Kb is the molal boiling point constant, and m is the molality of the solution.

ΔTb = 2.53°C/m x 0.0574 mol/kg = 0.145°C

Finally, we can calculate the boiling point of the solution:

boiling point = 80.1°C + 0.145°C = 80.245°C

Round to one decimal place: 80.2°C

The closest option is C) 80.9°C.

Question 9

What is the molar mass of a nonpolar molecular compound if 3.42 grams dissolved in 41.8 grams benzene begins to freeze at 1.17°C? The freezing point of pure benzene is 5.50°C and the molal freezing point constant,K_f,is -5.12°C/m.

Accepted Answer

First, we need to calculate the change in freezing point of the solution compared to pure benzene:

ΔTf = Tf,pure - Tf,soln = 5.50°C - 1.17°C = 4.33°C

Next, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

Assuming that all 3.42 grams of the compound dissolved in 41.8 grams of benzene, the mass of solvent is:

mass of solvent = 41.8 g / 1000 g/1 kg = 0.0418 kg

Using the freezing point depression equation:

ΔTf = KSf * molality

We can solve for molality:

molality = ΔTf / KSf = 4.33°C / (-5.12°C/m) = -0.846 m

Note that molality is negative because the freezing point depression occurs when solute is added to the solvent.

Now we can solve for the moles of solute:

moles of solute = molality * kg of solvent = -0.846 m * 0.0418 kg = -0.0354 mol

Again, the negative sign indicates that the compound is the limiting reagent and that there is an excess of benzene.

Finally, we can calculate the molar mass of the compound:

molar mass = mass of solute / moles of solute = 3.42 g / (-0.0354 mol) = 96.7 g/mol

Therefore, the correct answer is C) 96.7 g/mol.

Question 10

Maple syrup is made from the sap of the maple tree.When the sap is tapped from the maple tree it is 2.0% by mass sucrose.Maple syrup is 66% by mass sucrose.What mass of sap would be required to make 0.5 kg of maple syrup?

Accepted Answer

The answer of Maple syrup is made from the sap...

Question 11

A solution is prepared by dissolving 5.00 g of an unknown molecular solid in water to make 1.00 L of solution.The osmotic pressure of the solution is 1.61 atm at 25°C.What is the molar mass of the solute? (R = 0.0821 L.atm/mol.K)

Accepted Answer

To find the molar mass of the solute, we need to use the equation for osmotic pressure:

π = MRT

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

We know the osmotic pressure (1.61 atm), the temperature (25°C = 298 K), and we can calculate the molarity:

M = moles of solute / liters of solution

We are given that we have 5.00 g of solute in 1.00 L of solution, so we need to convert grams to moles:

moles of solute = 5.00 g / molar mass

Now we can substitute in the values we know:

1.61 atm = (5.00 g / molar mass) * (0.0821 L.atm/mol.K) * 298 K / 1.00 L

Solving for the molar mass gives:

molar mass = 5.00 g / (1.61 atm * 0.0821 L.atm/mol.K * 298 K) = 76.0 g/mol

Therefore, the answer is C.

Question 12

A solution is prepared by dissolving 4.78 g of an unknown nonelectrolyte in enough water to make 375 mL of solution.The osmotic pressure of the solution is 1.33 atm at 27°C.What is the molar mass of the solute? (R = 0.0821 L⋅atm/mol⋅K)

Accepted Answer

The molar mass can be calculated using the formula for osmotic pressure: π = (n/V)RT. Rearranging for molar mass (M), M = (mass * R * T) / (π * V). Substituting the given values, M = (4.78 g * 0.0821 L⋅atm/mol⋅K * 300 K) / (1.33 atm * 0.375 L) = 236 g/mol.

Quiz 10: Solutions

At 25°C,what is the osmotic pressure of a homogeneous solution consisting of 18.0 g urea (CON₂H₄) diluted with water to 3.00 L? (R = 0.0821 L.atm/mol.K)

An aqueous solution is composed of 7.50 g NaCl (MM = 58.44 g/mol) diluted to 0.100 L.Calculate the osmotic pressure of the solution at 298 K.(R = 0.0821 L.atm/mol.K)

Equal masses of water and ethylene glycol (C₂H₆O₂) are mixed.At what temperature will the mixture freeze? The molal freezing point constant for water is −1.86°C/m.

The molal freezing point constant for water is -1.86°C/m.At what temperature will a solution containing 8.27 g CaCl₂ and 45.0 g H₂O begin to freeze? Assume that no ion-pairing occurs between Ca²⁺ and Cl^-.

Which of the following electrolytes is likely to have a van't Hoff factor equal to 3?

A 0.230 m solution of an unknown electrolyte depresses the freezing point of water by 0.821°C.What is the Van't Hoff factor for this electrolyte? The freezing point depression constant for water is 1.86°C/m.

A solution is prepared by dissolving 4.21 g of a nonelectrolyte in 50.0 g of water.If the boiling point increases by 0.203°C,what is the molar mass of the solute? The boiling point elevation constant for water is 0.512°C/m.

What is the boiling point of a solution containing 0.80 g caffeine,C₈H₁₀N₄O₂,dissolved in 13.20 g benzene? The boiling point of pure benzene is 80.1°C and the molal boiling point constant,K_b,is 2.53°C/m.

What is the molar mass of a nonpolar molecular compound if 3.42 grams dissolved in 41.8 grams benzene begins to freeze at 1.17°C? The freezing point of pure benzene is 5.50°C and the molal freezing point constant,K_f,is -5.12°C/m.

Maple syrup is made from the sap of the maple tree.When the sap is tapped from the maple tree it is 2.0% by mass sucrose.Maple syrup is 66% by mass sucrose.What mass of sap would be required to make 0.5 kg of maple syrup?

A solution is prepared by dissolving 5.00 g of an unknown molecular solid in water to make 1.00 L of solution.The osmotic pressure of the solution is 1.61 atm at 25°C.What is the molar mass of the solute? (R = 0.0821 L.atm/mol.K)

A solution is prepared by dissolving 4.78 g of an unknown nonelectrolyte in enough water to make 375 mL of solution.The osmotic pressure of the solution is 1.33 atm at 27°C.What is the molar mass of the solute? (R = 0.0821 L⋅atm/mol⋅K)

Quiz 10: Solutions

At 25°C,what is the osmotic pressure of a homogeneous solution consisting of 18.0 g urea (CON2H4) diluted with water to 3.00 L? (R = 0.0821 L.atm/mol.K)

An aqueous solution is composed of 7.50 g NaCl (MM = 58.44 g/mol) diluted to 0.100 L.Calculate the osmotic pressure of the solution at 298 K.(R = 0.0821 L.atm/mol.K)

Equal masses of water and ethylene glycol (C2H6O2) are mixed.At what temperature will the mixture freeze? The molal freezing point constant for water is −1.86°C/m.

The molal freezing point constant for water is -1.86°C/m.At what temperature will a solution containing 8.27 g CaCl2 and 45.0 g H2O begin to freeze? Assume that no ion-pairing occurs between Ca2+ and Cl-.

Which of the following electrolytes is likely to have a van't Hoff factor equal to 3?

A 0.230 m solution of an unknown electrolyte depresses the freezing point of water by 0.821°C.What is the Van't Hoff factor for this electrolyte? The freezing point depression constant for water is 1.86°C/m.

A solution is prepared by dissolving 4.21 g of a nonelectrolyte in 50.0 g of water.If the boiling point increases by 0.203°C,what is the molar mass of the solute? The boiling point elevation constant for water is 0.512°C/m.

What is the boiling point of a solution containing 0.80 g caffeine,C8H10N4O2,dissolved in 13.20 g benzene? The boiling point of pure benzene is 80.1°C and the molal boiling point constant,Kb,is 2.53°C/m.

What is the molar mass of a nonpolar molecular compound if 3.42 grams dissolved in 41.8 grams benzene begins to freeze at 1.17°C? The freezing point of pure benzene is 5.50°C and the molal freezing point constant,Kf,is -5.12°C/m.

Maple syrup is made from the sap of the maple tree.When the sap is tapped from the maple tree it is 2.0% by mass sucrose.Maple syrup is 66% by mass sucrose.What mass of sap would be required to make 0.5 kg of maple syrup?

A solution is prepared by dissolving 5.00 g of an unknown molecular solid in water to make 1.00 L of solution.The osmotic pressure of the solution is 1.61 atm at 25°C.What is the molar mass of the solute? (R = 0.0821 L.atm/mol.K)

A solution is prepared by dissolving 4.78 g of an unknown nonelectrolyte in enough water to make 375 mL of solution.The osmotic pressure of the solution is 1.33 atm at 27°C.What is the molar mass of the solute? (R = 0.0821 L⋅atm/mol⋅K)

At 25°C,what is the osmotic pressure of a homogeneous solution consisting of 18.0 g urea (CON₂H₄) diluted with water to 3.00 L? (R = 0.0821 L.atm/mol.K)

Equal masses of water and ethylene glycol (C₂H₆O₂) are mixed.At what temperature will the mixture freeze? The molal freezing point constant for water is −1.86°C/m.

The molal freezing point constant for water is -1.86°C/m.At what temperature will a solution containing 8.27 g CaCl₂ and 45.0 g H₂O begin to freeze? Assume that no ion-pairing occurs between Ca²⁺ and Cl^-.

What is the boiling point of a solution containing 0.80 g caffeine,C₈H₁₀N₄O₂,dissolved in 13.20 g benzene? The boiling point of pure benzene is 80.1°C and the molal boiling point constant,K_b,is 2.53°C/m.

What is the molar mass of a nonpolar molecular compound if 3.42 grams dissolved in 41.8 grams benzene begins to freeze at 1.17°C? The freezing point of pure benzene is 5.50°C and the molal freezing point constant,K_f,is -5.12°C/m.