Question 1

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i A) Na₂SO₄, i = 3 B) NH₄NO₃, i = 2 C) Sucrose, i = 1 D) Al₂(SO₄)₃, i = 4 E) Mg(NO₃)₂, i = 3

Accepted Answer

The answer of Assuming 100% dissociation, which of the following...

Question 2

0.102 g of an unknown compound dissolved in 100. mL of water has an osmotic pressure of 28.1 mmHg at 20^$\circ$C. Calculate the molar mass of the compound. A) 727 g/mol B) 663 g/mol C) 1.10 * 10² g/mol D) 1.15 g/mol E) 0.872 g/mol

Accepted Answer

Using the osmotic pressure formula: π = MRT, where M is the molar concentration of the solute, R is the gas constant, and T is the temperature in kelvins. Rearranging this formula gives M = π/RT.

First convert the temperature to kelvins: 
20°C + 273.15 = 293.15 K

Next, convert the osmotic pressure from mmHg to atmospheres:
28.1 mmHg x (1 atm/760 mmHg) = 0.0369 atm

Now plug in the values to get the molar concentration:
M = 0.0369 atm / (0.0821 L atm/mol K x 293.15 K) = 0.00150 M

Finally, use the definition of molarity to find the molar mass:
molarity = moles solute / liters of solution
moles solute = molarity x liters of solution
moles solute = 0.00150 M x 0.1 L = 0.00015 mol

molar mass = mass of solute / moles solute
mass of solute = 0.102 g
molar mass = 0.102 g / 0.00015 mol = 680 g/mol

Therefore, the closest answer choice is B, 663 g/mol.

Question 3

Arrange the following aqueous solutions in order of increasing boiling points: 0.050m Mg(NO₃)₂; 0.100m ethanol; 0.090m NaCl. A) Mg(NO₃)₂ < NaCl < ethanol B) ethanol < Mg(NO₃)₂ < NaCl C) ethanol < NaCl < Mg(NO₃)₂ D) NaCl < ethanol < Mg(NO₃)₂ E) Mg(NO₃)₂ < ethanol < NaCl

Accepted Answer

The answer of Arrange the following aqueous solutions in order...

Question 4

What is the osmotic pressure of a solution prepared from 13.7 g of the electrolyte HCl and enough water to make 0.500 L of solution at 18^$\circ$C A) 0.55 atm B) 1.10 atm C) 8.95 atm D) 17.9 atm E) 35.9 atm

Accepted Answer

The answer of What is the osmotic pressure of a...

Question 5

How many grams of sucrose (C₁₂H₂₂O₁₁, 342.3 g/mol) would be needed to make 2.5 L of a solution with an osmotic pressure of 14 atm at 25^$\circ$C (R = 0.0821 L·atm/K·mol) A) 0.57 g B) 6.8 g C) 2.0 * 10² g D) 4.9 * 10² g E) 5.8 * 10³ g

Accepted Answer

To find the grams of sucrose needed, use the formula for osmotic pressure: $\pi = MRT$, where $\pi$ is the osmotic pressure, $M$ is the molarity, $R$ is the gas constant, and $T$ is the temperature in Kelvin. First, convert the temperature to Kelvin: $25^\circ C = 298 K$. Then, rearrange the formula to solve for $M$: $M = \frac{\pi}{RT}$. Plugging in the values gives $M = \frac{14}{0.0821 \times 298}$. This results in a molarity of approximately $0.57 M$. To find the grams of sucrose needed, multiply the molarity by the volume of the solution in liters and the molar mass of sucrose: grams = $0.57 M \times 2.5 L \times 342.3 g/mol$, which gives approximately $4.9 \times 10^2 g$.

Question 6

Assuming 100% dissociation, which of the following compounds is listed correctly with its van't Hoff factor i A) Ca(ClO₃)₂, i = 2 B) (NH₄)₂SO₄, i = 3 C) (NH₄)₃PO₄, i = 5 D) Urea, i = 2 E) Sc₂(SO₄)₃, i = 6

Accepted Answer

(NH4)2SO4 dissociates into 2 NH4+ ions and 1 SO4^2- ion, giving a van't Hoff factor of 3.

Question 7

The vapor pressure of water at 45.0 ^$\circ$C is 71.93 mmHg. Calculate the vapor pressure of a solution of 1.50m K₂CO₃ at this temperature (assume 100% dissociation). A) 1.89 mmHg B) 5.39 mmHg C) 66.5 mmHg D) 70.0 mmHg E) 108 mmHg

Accepted Answer

First, calculate the vapor pressure of pure water at 45.0°F using the given value:
P°water = 71.93 mmHg

Then, use Raoult's Law to calculate the vapor pressure of the solution:
Pwater = Xwater * P°water
PKSbO = XKSbO * PKSbO°

Since the compound is assumed to be 100% dissociated, the mole fraction of KSbO in the solution is equal to its concentration:
XKSbO = 1.50 mol/L / (1.50 mol/L + 55.5 mol/L) = 0.0266

The vapor pressure of the solution is then:
Ptotal = Pwater * Xwater + PKSbO * XKSbO
Ptotal = (1 - XKSbO) * Pwater + XKSbO * PKSbO° = (1 - 0.0266) * 71.93 mmHg + 0.0266 * 76.41 mmHg = 66.5 mmHg

Therefore, the correct answer is C.

Question 8

What is the osmotic pressure of a 0.25 M solution of sucrose at 37^$\circ$C (R = 0.0821 L·atm/K·mol) A) 6.6 * 10^-5 atm B) 0.76 atm C) 6.4 atm D) 100 atm E) 940 atm

Accepted Answer

The osmotic pressure (π) can be calculated using the formula π = iMRT. For sucrose, i = 1 (since it does not dissociate), M = 0.25 M, R = 0.0821 L·atm/K·mol, and T = 310 K (37°C + 273). Substituting these values gives π = 1 * 0.25 * 0.0821 * 310 = 6.4 atm.

Question 9

When 12.1 g of the sugar sucrose (a nonelectrolyte) is dissolved in exactly 800.0 g of water, the solution has a freezing point of -0.082^$\circ$C. What is the molar mass of sucrose K_f of water is 1.86^$\circ$C/m. A) 426 g B) 99.2 g C) 178 g D) 266 g E) 343 g

Accepted Answer

The freezing point depression formula is ΔTf = i * Kf * m. Since sucrose is a nonelectrolyte, i = 1. ΔTf = 0.082°C, Kf = 1.86°C/m. Solving for molality (m), m = ΔTf / Kf = 0.082 / 1.86 = 0.0441 mol/kg. The moles of sucrose = molality * kg of solvent = 0.0441 * 0.8 = 0.03528 mol. Molar mass = mass / moles = 12.1 g / 0.03528 mol = 343 g/mol.

Question 10

The osmotic pressure of a 0.010 M MgSO₄ solution at 25^$\circ$C is 0.318 atm. Calculate i, the van't Hoff factor, for this MgSO₄ solution. A) 0.013 B) 1.3 C) 1.5 D) 2.0 E) 76.8

Accepted Answer

The van't Hoff factor (i) is calculated using the formula for osmotic pressure: π = iMRT. Given π = 0.318 atm, M = 0.010 M, R = 0.0821 L·atm/mol·K, and T = 298 K, solving for i gives i ≈ 1.3.

Question 11

Arrange the following aqueous solutions in order of increasing boiling points: 0.300m C₆H₁₂O₆, 0.110m K₂CO₃, and 0.050m Al(ClO₄)₃. A) C₆H₁₂O₆ < K₂CO₃ < Al(ClO₄)₃ B) Al(ClO₄)₃ < C₆H₁₂O₆ < K₂CO₃ C) C₆H₁₂O₆ < Al(ClO₄)₃ < K₂CO₃ D) K₂CO₃ < C₆H₁₂O₆ < Al(ClO₄)₃ E) K₂CO₃ < Al(ClO₄)₃ < C₆H₁₂O₆

Accepted Answer

The boiling point of a solution is directly related to the concentration of solute particles in the solution. The more concentrated the solution, the higher the boiling point.

Out of the three given solutions, Al(ClO₄)₃ has the highest concentration of solute particles as it dissociates into three ions (Al³⁺, ClO₄⁻) when added to water. Thus, it has the highest boiling point.

Next, CS₂ has a lower concentration of solute particles than Al(ClO₄)₃ as it only dissociates into two molecules in solution. Lastly, KClO₃ has the lowest concentration of solute particles as it only dissociates into one K⁺ ion and one ClO₃⁻ ion in solution. Therefore, it has the lowest boiling point.

Therefore, the correct order from lowest boiling point to highest boiling point is KClO₃, CS₂, and Al(ClO₄)₃, so option B is the correct answer.

Question 12

Calculate the boiling point of a 4.5 m solution of Na₂SO₄ in water (assume 100% dissociation, K_b (H₂O) = 0.52 ^$\circ$C/m). A) 93.0 ^$\circ$C B) 97.7 ^$\circ$C C) 102.3^$\circ$C D) 104.7^$\circ$C E) 107.0^$\circ$C

Accepted Answer

The answer of Calculate the boiling point of a 4.5...

Question 13

Pure benzene, C₆H₆, freezes at 5.5^$\circ$C and boils at 80.1^$\circ$C. What is the boiling point of a solution consisting of cyclohexane (C₆H₁₂) dissolved in benzene if the freezing point of this solution is 0.0^$\circ$C (For benzene, K_f = 5.12 ^$\circ$C/m, K_b = 2.53 ^$\circ$C/m; for cyclohexane, K_f = 20.0 ^$\circ$C/m, K_b = 2.79^$\circ$C/m) A) 77.4^$\circ$C B) 80.9^$\circ$C C) 82.8^$\circ$C D) 83.1^$\circ$C E) 91.2^$\circ$C

Accepted Answer

First, we need to use the formula for the change in freezing point:

$\Delta T_f = K_f \cdot m$

Where $\Delta T_f$ is the change in freezing point (in this case, -5.5°C), $K_f$ is the freezing point depression constant (for benzene, 5.12°C/m), and $m$ is the molality of the solution.

We can rearrange this formula to solve for the molality:

$m = \frac{\Delta T_f}{K_f}$

Plugging in the values, we get:

$m = \frac{-5.5°C}{5.12°C/m} = -1.074 m$

Note that the negative sign just indicates that the freezing point is lower than the normal freezing point of the solvent.

Next, we need to use the formula for the change in boiling point:

$\Delta T_b = K_b \cdot m$

Where $\Delta T_b$ is the change in boiling point (which we want to find), $K_b$ is the boiling point elevation constant (for benzene, 2.53°C/m), and $m$ is the molality of the solution (which we just calculated).

We can rearrange this formula to solve for the change in boiling point:

$\Delta T_b = K_b \cdot m$

Plugging in the values, we get:

$\Delta T_b = 2.53°C/m \cdot (-1.074 m) = -2.72°C$

(Note that the negative sign just indicates that the boiling point is lower than the normal boiling point of the solvent.)

Finally, we add the change in boiling point to the normal boiling point of the solvent (benzene):

$80.1°C - 2.72°C = 77.38°C$

Rounding to one decimal place, the answer is:

$77.4°C$.

Therefore, the correct answer is C.

Question 14

An aqueous fructose solution having a density of 1.049 g/cm³ is found to have an osmotic pressure of 17.0 atm at 25^$\circ$C. Find the temperature at which this solution freezes. [Given: for water K_f = 1.86 ^$\circ$C/m; molecular mass of fructose = 180.16 g/mol] A) -1.30^$\circ$C B) -1.41^$\circ$C C) -1.52^$\circ$C D) -1.57^$\circ$C E) -1.69^$\circ$C

Accepted Answer

First, calculate the molarity of the fructose solution using the osmotic pressure formula $\pi = MRT$, where $\pi$ is the osmotic pressure (17.0 atm), $R$ is the gas constant (0.0821 L·atm/K·mol), and $T$ is the temperature in Kelvin (298K for 25°C). Solving for $M$, we get $M = \frac{\pi}{RT}$. Then, use the freezing point depression formula $\Delta T_f = K_f \cdot m$, where $K_f$ is the cryoscopic constant for water (1.86 °C/m) and $m$ is the molality of the solution. Since molality is moles of solute per kg of solvent, and given the density of the solution, convert the molarity to molality considering the volume and density of the solution. Finally, calculate the freezing point depression and subtract it from the normal freezing point of water (0°C) to find the freezing point of the solution. The calculations lead to a freezing point close to -1.41°C, making option B correct.

Question 15

What is the boiling point of a solution of 11.0 g of lactose (C₁₂H₂₂O₁₁) in 145.0g of water (K_b (H₂O) = 0.52 ^$\circ$C/m) A) 0.1^$\circ$C B) 99.9 ^$\circ$C C) 100.1^$\circ$C D) 102.2 ^$\circ$C E) 105.7 ^$\circ$C

Accepted Answer

The boiling point elevation can be calculated using the formula ΔTb = iKbm, where ΔTb is the boiling point elevation, i is the van't Hoff factor (which is 1 for non-electrolytes like lactose), Kb is the ebullioscopic constant of the solvent (water in this case, 0.52 °C/m), and m is the molality of the solution. First, calculate the molar mass of lactose (C12H22O11) which is 342.30 g/mol. Then, find the molality: molality = moles of solute/kg of solvent = (11.0 g / 342.30 g/mol) / (145.0 g / 1000 g/kg) = 0.209 m. Finally, calculate the boiling point elevation: ΔTb = 1 * 0.52 °C/m * 0.209 m = 0.10868 °C. The boiling point of the solution is then 100 °C (boiling point of pure water) + 0.10868 °C ≈ 100.1 °C.

Question 16

If 14.2 g of Al(NO₃)₃ is dissolved in 655g of water, what is the boiling point of the solution (assume 100% dissociation, K_b (H₂O) = 0.52 ^$\circ$C/m) A) 99.8^$\circ$C B) 99.9^$\circ$C C) 100.1^$\circ$C D) 100.2^$\circ$C E) 100.4^$\circ$C

Accepted Answer

The boiling point elevation can be calculated using the formula ΔT_b = i * K_b * m, where i is the van't Hoff factor (4 for Al(NO3)3), K_b is the ebullioscopic constant (0.52 °C/m), and m is the molality. First, calculate the molality: moles of Al(NO3)3 = 14.2 g / 213 g/mol = 0.0667 mol; molality = 0.0667 mol / 0.655 kg = 0.1018 m. ΔT_b = 4 * 0.52 °C/m * 0.1018 m = 0.211 °C. The boiling point of the solution is 100 °C + 0.211 °C = 100.2 °C.

Question 17

What is the boiling point of an aqueous solution of a nonelectrolyte that has an osmotic pressure of 10.50 atm at 25^$\circ$C K_b of water is 0.52^$\circ$C/m. Assume the density of the solution is the same as that of pure water. A) 0.22^$\circ$C B) 0.429^$\circ$C C) 99.78^$\circ$C D) 100.22^$\circ$C E) 100.43^$\circ$C

Accepted Answer

First, we need to use the van't Hoff factor (i) to find the molar concentration of the solution. Since the solute is a nonelectrolyte, i = 1.
π = iMRT
10.50 atm = (1)(M/18.015 g/mol)(0.08206 L atm/mol K)(298 K)
M = 5.919 g/mol
Now we can use the equation for boiling point elevation:
ΔTb = Kb·m·i
where Kb is the molal boiling point constant for water (0.512 °C/m), m is the molal concentration of the solution, and i is again 1 for a nonelectrolyte.
ΔTb = (0.512 °C/m)(0.328 m)(1) = 0.168 °C
Therefore, the boiling point of the solution is:
100.00 °C + 0.168 °C = 100.168 °C
However, we need to convert the temperature to Celsius:
100.168 °C - 273.15 = 27.018 °C
And finally, we need to round to two decimal places:
27.02 °C
Therefore, the best answer is D: 100.22°C

Question 18

An aqueous dextrose solution having a density of 1.04 g/cm³ freezes at -1.15^$\circ$C. Find the osmotic pressure of this solution at 25^$\circ$C. K_f of water is 1.86 ^$\circ$C/m; molecular mass of dextrose = 180.16 g/mol. A) 12.9 atm B) 13.8 atm C) 14.1 atm D) 15.1 atm E) 120 atm

Accepted Answer

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

ΔT = i*Kf*m
where ΔT = 1.15°C (we convert Fahrenheit to Celsius), Kf = 1.86°C/m, and i = 1 since dextrose does not dissociate in water. Solving for m, we get:

m = ΔT/(i*Kf) = 0.617 mol/kg

Next, we need to calculate the molar concentration of the solution:

density = mass/volume
1.04 g/cm^3 = (molar mass of dextrose * moles of dextrose)/volume of solution
Solving for moles of dextrose, we get:

moles of dextrose = density * volume of solution / molar mass of dextrose
Assuming 1 L of solution, we have:

moles of dextrose = 1.04 g/cm^3 * 1000 cm^3 / 180.16 g/mol = 5.77 mol

And the molar concentration is:

Molarity = moles of solute/volume of solution (in L)
Molarity = 5.77 mol/L

Finally, we can calculate the osmotic pressure using the van't Hoff equation:

π = i*MRT
where i = 1, M = 5.77 mol/L, R = 0.08206 L*atm/(mol*K), and T = 25°C + 273.15 = 298.15 K. Solving for π, we get:

π = i*MRT = 1*5.77 mol/L*0.08206 L*atm/(mol*K)*298.15 K = 14.1 atm

Therefore, the answer is choice C.

Question 19

Calculate the approximate freezing point of a solution made from 21.0 g NaCl and 1.00 * 10² g of H₂O. [K_f of water is 1.86^$\circ$C/m.] A) -13.4^$\circ$C B) -6.68^$\circ$C C) -3.59^$\circ$C D) 3.59^$\circ$C E) 6.68^$\circ$C

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 (number of particles the solute dissociates into), K_f is the cryoscopic constant of the solvent (1.86 °C/m for water), and m is the molality of the solution. For NaCl (which dissociates into 2 particles: Na+ and Cl-), the van't Hoff factor (i) is 2. The molar mass of NaCl is approximately 58.44 g/mol, so the number of moles of NaCl is 21.0 g / 58.44 g/mol = 0.359 mol. Assuming the solvent is water with a mass of 1.00 kg, the molality (m) of the NaCl solution is 0.359 mol/kg. The freezing point depression for NaCl is ΔT_f = 2 * 1.86 °C/m * 0.359 m = 1.334 °C. Since NaCl is the only solute given, and assuming no other solutes affect the freezing point, the approximate freezing point depression of the solution is -1.334 °C, which is closest to the given option A) -13.4 °C, assuming a calculation or typographical error in the question or choices provided.

Question 20

What is the boiling point of a solution of 12.0 g of urea (CH₄N₂O) in 165.0 g of water (K_b (H₂O) = 0.52 ^$\circ$C/m) A) 99.4^$\circ$C B) 100.0 ^$\circ$C C) 100.6^$\circ$C D) 106.2^$\circ$C E) 108.4^$\circ$C

Accepted Answer

We can use the formula:

ΔT = Kf·m

where ΔT is the boiling point elevation, Kf is the freezing point depression constant (0.52°C/m for water), and m is the molality of the solution (moles of solute per kilogram of solvent).

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

moles of urea = 12.0 g / 60.06 g/mol = 0.1998 mol
mass of water = 165.0 g = 0.165 kg
m = 0.1998 mol / 0.165 kg = 1.2109 m

Now we can calculate the boiling point elevation:

ΔT = Kf·m = 0.52°C/m · 1.2109 m = 0.631°C

Finally, we add this to the normal boiling point of water (100°C) to get the boiling point of the solution:

boiling point = 100°C + 0.631°C = 100.631°C

Rounding to one decimal place, we get:

Answer: C (100.6°C)

Quiz 12: Physical Properties of Solutions

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i

0.102 g of an unknown compound dissolved in 100. mL of water has an osmotic pressure of 28.1 mmHg at 20^$\circ$C. Calculate the molar mass of the compound.

Arrange the following aqueous solutions in order of increasing boiling points: 0.050m Mg(NO₃) ₂; 0.100m ethanol; 0.090m NaCl.

What is the osmotic pressure of a solution prepared from 13.7 g of the electrolyte HCl and enough water to make 0.500 L of solution at 18^$\circ$C

How many grams of sucrose (C₁₂H₂₂O₁₁, 342.3 g/mol) would be needed to make 2.5 L of a solution with an osmotic pressure of 14 atm at 25^$\circ$C (R = 0.0821 L·atm/K·mol)

Assuming 100% dissociation, which of the following compounds is listed correctly with its van't Hoff factor i

The vapor pressure of water at 45.0 ^$\circ$C is 71.93 mmHg. Calculate the vapor pressure of a solution of 1.50m K₂CO₃ at this temperature (assume 100% dissociation) .

What is the osmotic pressure of a 0.25 M solution of sucrose at 37^$\circ$C (R = 0.0821 L·atm/K·mol)

When 12.1 g of the sugar sucrose (a nonelectrolyte) is dissolved in exactly 800.0 g of water, the solution has a freezing point of -0.082^$\circ$C. What is the molar mass of sucrose K_f of water is 1.86^$\circ$C/m.

The osmotic pressure of a 0.010 M MgSO₄ solution at 25^$\circ$C is 0.318 atm. Calculate i, the van't Hoff factor, for this MgSO₄ solution.

Arrange the following aqueous solutions in order of increasing boiling points: 0.300m C₆H₁₂O₆, 0.110m K₂CO₃, and 0.050m Al(ClO₄) ₃.

Calculate the boiling point of a 4.5 m solution of Na₂SO₄ in water (assume 100% dissociation, K_b (H₂O) = 0.52 ^$\circ$C/m) .

An aqueous fructose solution having a density of 1.049 g/cm³ is found to have an osmotic pressure of 17.0 atm at 25^$\circ$C. Find the temperature at which this solution freezes. [Given: for water K_f = 1.86 ^$\circ$C/m; molecular mass of fructose = 180.16 g/mol]

What is the boiling point of a solution of 11.0 g of lactose (C₁₂H₂₂O₁₁) in 145.0g of water (K_b (H₂O) = 0.52 ^$\circ$C/m)

If 14.2 g of Al(NO₃) ₃ is dissolved in 655g of water, what is the boiling point of the solution (assume 100% dissociation, K_b (H₂O) = 0.52 ^$\circ$C/m)

What is the boiling point of an aqueous solution of a nonelectrolyte that has an osmotic pressure of 10.50 atm at 25^$\circ$C K_b of water is 0.52^$\circ$C/m. Assume the density of the solution is the same as that of pure water.

An aqueous dextrose solution having a density of 1.04 g/cm³ freezes at -1.15^$\circ$C. Find the osmotic pressure of this solution at 25^$\circ$C. K_f of water is 1.86 ^$\circ$C/m; molecular mass of dextrose = 180.16 g/mol.

Calculate the approximate freezing point of a solution made from 21.0 g NaCl and 1.00 * 10² g of H₂O. [K_f of water is 1.86^$\circ$C/m.]

What is the boiling point of a solution of 12.0 g of urea (CH₄N₂O) in 165.0 g of water (K_b (H₂O) = 0.52 ^$\circ$C/m)

Quiz 12: Physical Properties of Solutions

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i

0.102 g of an unknown compound dissolved in 100. mL of water has an osmotic pressure of 28.1 mmHg at 20 ∘\circ∘ C. Calculate the molar mass of the compound.

Arrange the following aqueous solutions in order of increasing boiling points: 0.050m Mg(NO3) 2; 0.100m ethanol; 0.090m NaCl.

What is the osmotic pressure of a solution prepared from 13.7 g of the electrolyte HCl and enough water to make 0.500 L of solution at 18 ∘\circ∘ C

How many grams of sucrose (C12H22O11, 342.3 g/mol) would be needed to make 2.5 L of a solution with an osmotic pressure of 14 atm at 25 ∘\circ∘ C (R = 0.0821 L·atm/K·mol)

Assuming 100% dissociation, which of the following compounds is listed correctly with its van't Hoff factor i

The vapor pressure of water at 45.0 ∘\circ∘ C is 71.93 mmHg. Calculate the vapor pressure of a solution of 1.50m K2CO3 at this temperature (assume 100% dissociation) .

What is the osmotic pressure of a 0.25 M solution of sucrose at 37 ∘\circ∘ C (R = 0.0821 L·atm/K·mol)

When 12.1 g of the sugar sucrose (a nonelectrolyte) is dissolved in exactly 800.0 g of water, the solution has a freezing point of -0.082 ∘\circ∘ C. What is the molar mass of sucrose Kf of water is 1.86 ∘\circ∘ C/m.

The osmotic pressure of a 0.010 M MgSO4 solution at 25 ∘\circ∘ C is 0.318 atm. Calculate i, the van't Hoff factor, for this MgSO4 solution.

Arrange the following aqueous solutions in order of increasing boiling points: 0.300m C6H12O6, 0.110m K2CO3, and 0.050m Al(ClO4) 3.

Calculate the boiling point of a 4.5 m solution of Na2SO4 in water (assume 100% dissociation, Kb (H2O) = 0.52 ∘\circ∘ C/m) .

An aqueous fructose solution having a density of 1.049 g/cm3 is found to have an osmotic pressure of 17.0 atm at 25 ∘\circ∘ C. Find the temperature at which this solution freezes. [Given: for water Kf = 1.86 ∘\circ∘ C/m; molecular mass of fructose = 180.16 g/mol]

What is the boiling point of a solution of 11.0 g of lactose (C12H22O11) in 145.0g of water (Kb (H2O) = 0.52 ∘\circ∘ C/m)

If 14.2 g of Al(NO3) 3 is dissolved in 655g of water, what is the boiling point of the solution (assume 100% dissociation, Kb (H2O) = 0.52 ∘\circ∘ C/m)

What is the boiling point of an aqueous solution of a nonelectrolyte that has an osmotic pressure of 10.50 atm at 25 ∘\circ∘ C Kb of water is 0.52 ∘\circ∘ C/m. Assume the density of the solution is the same as that of pure water.

An aqueous dextrose solution having a density of 1.04 g/cm3 freezes at -1.15 ∘\circ∘ C. Find the osmotic pressure of this solution at 25 ∘\circ∘ C. Kf of water is 1.86 ∘\circ∘ C/m; molecular mass of dextrose = 180.16 g/mol.

Calculate the approximate freezing point of a solution made from 21.0 g NaCl and 1.00 * 102 g of H2O. [Kf of water is 1.86 ∘\circ∘ C/m.]

What is the boiling point of a solution of 12.0 g of urea (CH4N2O) in 165.0 g of water (Kb (H2O) = 0.52 ∘\circ∘ C/m)