Assuming 100% dissociation

Assuming 100% dissociation, calculate the freezing point (f) and boiling point (?b) of 3.29 ? AgNO3(aq).

The correct answer and explanation is :

To determine the freezing point (Tf) and boiling point (Tb) of a 3.29 m AgNO₃(aq) solution, assuming 100% dissociation, we can apply the principles of colligative properties. These properties depend on the number of solute particles in a given amount of solvent, irrespective of the nature of the solute particles.

Step 1: Understanding the Dissociation of AgNO₃

Silver nitrate (AgNO₃) dissociates completely in water into silver ions (Ag⁺) and nitrate ions (NO₃⁻):

[ \text{AgNO₃} \rightarrow \text{Ag}^+ + \text{NO}_3^- ]

This dissociation results in two particles per formula unit, giving a van’t Hoff factor (i) of 2.

Step 2: Calculating Freezing Point Depression (ΔTf)

The freezing point depression is given by the formula:

[ \Delta T_f = i \times K_f \times m ]

Where:

  • ( i ) is the van’t Hoff factor (2 for AgNO₃),
  • ( K_f ) is the cryoscopic constant (freezing point depression constant) for water, approximately 1.86 °C·kg/mol,
  • ( m ) is the molality of the solution (3.29 mol/kg).

Plugging in the values:

[ \Delta T_f = 2 \times 1.86\,^\circ\text{C} \cdot \text{kg/mol} \times 3.29\,\text{mol/kg} ]
[ \Delta T_f = 12.22\,^\circ\text{C} ]

Therefore, the freezing point of the solution is:

[ T_f = 0\,^\circ\text{C} – 12.22\,^\circ\text{C} = -12.22\,^\circ\text{C} ]

Step 3: Calculating Boiling Point Elevation (ΔTb)

The boiling point elevation is calculated using:

[ \Delta T_b = i \times K_b \times m ]

Where:

  • ( K_b ) is the ebullioscopic constant (boiling point elevation constant) for water, approximately 0.512 °C·kg/mol.

Calculating:

[ \Delta T_b = 2 \times 0.512\,^\circ\text{C} \cdot \text{kg/mol} \times 3.29\,\text{mol/kg} ]
[ \Delta T_b = 3.37\,^\circ\text{C} ]

Thus, the boiling point of the solution is:

[ T_b = 100\,^\circ\text{C} + 3.37\,^\circ\text{C} = 103.37\,^\circ\text{C} ]

Conclusion

For a 3.29 m AgNO₃(aq) solution, assuming complete dissociation:

  • The freezing point is approximately -12.22 °C.
  • The boiling point is approximately 103.37 °C.

These changes are due to the presence of solute particles disrupting the formation of the solid lattice in freezing and affecting the vapor pressure in boiling.

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