An aqueous solution of urea has a freezing point of -0.52°C. Predict the osmotic pressure of the solution at 37°C [K_f = 1.86, assuming that the molar concentration and molality are numerically equal].

Question

Prepp Team · Accepted Answer

Understanding Colligative Properties: Freezing Point Depression and Osmotic Pressure

This question involves two important colligative properties of solutions: freezing point depression and osmotic pressure. Colligative properties depend only on the number of solute particles in a solution, not on their identity. We are given information about the freezing point of a urea solution and asked to predict its osmotic pressure at a different temperature.

Analyzing the Given Information

Freezing point of the urea solution ($T_f$) = -0.52°C
Normal freezing point of water ($T_f^0$) = 0°C (since it's an aqueous solution)
Freezing point depression constant for water ($K_f$) = 1.86 °C kg/mol
Temperature for predicting osmotic pressure ($T$) = 37°C
Assumption: Molar concentration (M) is numerically equal to molality (m).
Urea is a non-electrolyte, meaning it does not dissociate in water. Therefore, the van't Hoff factor ($i$) for urea is 1.

Step-by-Step Calculation of Osmotic Pressure

Step 1: Calculate the Freezing Point Depression ($\Delta T_f$)

The freezing point depression is the difference between the normal freezing point of the solvent (water) and the freezing point of the solution.

$$ \Delta T_f = T_f^0 - T_f $$ $$ \Delta T_f = 0°C - (-0.52°C) $$ $$ \Delta T_f = 0.52°C $$

Step 2: Determine the Molality (m) of the Urea Solution

The freezing point depression is related to the molality of the solution by the formula:

$$ \Delta T_f = i \times K_f \times m $$

Since urea is a non-electrolyte, $i=1$. We can rearrange the formula to solve for molality ($m$):

$$ m = \frac{\Delta T_f}{i \times K_f} $$ $$ m = \frac{0.52°C}{1 \times 1.86 \, \text{°C kg/mol}} $$ $$ m \approx 0.2796 \, \text{mol/kg} $$

The molality of the urea solution is approximately 0.2796 mol/kg.

Step 3: Determine the Molar Concentration (M)

The question states that we can assume the molar concentration (M) is numerically equal to the molality (m). This assumption is generally valid for dilute aqueous solutions.

$$ M \approx m $$ $$ M \approx 0.2796 \, \text{mol/L} $$

The molar concentration of the urea solution is approximately 0.2796 M.

Step 4: Convert the Temperature to Kelvin

The osmotic pressure calculation requires the temperature to be in Kelvin.

$$ T(K) = T(°C) + 273.15 $$ $$ T(K) = 37 + 273.15 $$ $$ T(K) = 310.15 \, K $$

Step 5: Calculate the Osmotic Pressure ($\Pi$)

The osmotic pressure is related to the molar concentration by the formula:

$$ \Pi = i \times M \times R \times T $$

Where:

$\Pi$ is the osmotic pressure
$i$ is the van't Hoff factor (1 for urea)
$M$ is the molar concentration (approx. 0.2796 mol/L)
$R$ is the ideal gas constant. We use the value $R = 0.0821 \, \text{L atm / (mol K)}$ to get the pressure in atmospheres.
$T$ is the temperature in Kelvin (310.15 K)

Substitute the values into the formula:

$$ \Pi = 1 \times 0.2796 \, \text{mol/L} \times 0.0821 \, \text{L atm / (mol K)} \times 310.15 \, K $$ $$ \Pi \approx 7.11 \, \text{atm} $$

The calculated osmotic pressure of the urea solution at 37°C is approximately 7.11 atm. This value is closest to the option 7.1 atm.

Summary of Results

Property	Value
Freezing Point Depression ($\Delta T_f$)	0.52 °C
Molality (m)	~0.2796 mol/kg
Molar Concentration (M)	~0.2796 mol/L
Temperature (T)	310.15 K
Osmotic Pressure ($\Pi$)	~7.11 atm

Revision Table: Colligative Property Formulas

Colligative Property	Formula (for non-electrolytes, i=1)	Variables
Freezing Point Depression	$\Delta T_f = K_f \times m$	$\Delta T_f$: freezing point depression, $K_f$: molal freezing point depression constant, $m$: molality
Boiling Point Elevation	$\Delta T_b = K_b \times m$	$\Delta T_b$: boiling point elevation, $K_b$: molal boiling point elevation constant, $m$: molality
Osmotic Pressure	$\Pi = M \times R \times T$	$\Pi$: osmotic pressure, $M$: molar concentration, $R$: ideal gas constant, $T$: absolute temperature
Relative Lowering of Vapor Pressure	$\frac{\Delta P}{P^0} = x_{\text{solute}}$	$\Delta P$: vapor pressure lowering, $P^0$: vapor pressure of pure solvent, $x_{\text{solute}}$: mole fraction of solute

Additional Information on Colligative Properties and Urea Solution

Colligative properties are a cornerstone concept in solution chemistry, explaining how adding a solute affects the physical properties of a solvent. Understanding these properties is crucial for various applications, including determining the molar mass of unknown substances, designing biological systems, and understanding processes like osmosis in living cells.

Urea as a Solute: Urea ($CO(NH_2)_2$) is a simple organic molecule that is highly soluble in water. It is a non-electrolyte because it does not dissociate into ions when dissolved in water. This makes calculations involving colligative properties simpler as the van't Hoff factor ($i$) is equal to 1. Electrolytes, like salts (e.g., NaCl), dissociate into multiple ions, so their van't Hoff factor would be greater than 1, reflecting the total number of particles formed per formula unit. For instance, NaCl dissociates into Na$^+$ and Cl$^-$ ions, giving an ideal $i=2$.

Molality vs. Molarity Assumption: Molality (moles of solute per kilogram of solvent) and molarity (moles of solute per liter of solution) are different concentration units. The assumption that they are numerically equal simplifies calculations but is only strictly accurate for very dilute aqueous solutions where the volume of the solution is approximately equal to the mass of the solvent (since the density of water is approximately 1 kg/L). In more concentrated solutions, the difference between molality and molarity becomes significant.

Ideal Gas Constant (R): The value of R used in the osmotic pressure formula depends on the units of pressure and volume. When pressure is in atmospheres (atm) and volume in liters (L), the appropriate value is $R = 0.0821 \, \text{L atm / (mol K)}$. If pressure were in Pascals (Pa) and volume in cubic meters ($m^3$), a different R value ($8.314 \, \text{J / (mol K)}$) would be used.

Prepp Team · Answer

1 atm

Prepp Team · Answer

11 atm

Prepp Team · Answer

8.5 atm