Question

80 mole percent of MgCl₂ is dissociated in aqueous solution. The vapour pressure of 1.0 molal aqueous solution of M_gCl₂ at 38°C is ____mm Hg. (Nearest integer)
Given : Vapour pressure of water at 38°C is 50 mm Hg.

Answer 1

Correct Answer: 48

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<b>Solution:</b>

1.  Calculate the van't Hoff factor (i):

    MgCl2 dissociates into Mg²⁺ and 2Cl^-, so it dissociates into 3 ions.
    The dissociation reaction is: MgCl2 → Mg²⁺ + 2Cl^-
    Degree of dissociation (α) = 80% = 0.8
    i = 1 + α(n - 1), where n is the number of ions produced.
    i = 1 + 0.8 (3 - 1) = 1 + 0.8 * 2 = 1 + 1.6 = 2.6

2.  Calculate the effective molality:

   Effective molality = i * molality
   Effective molality = 2.6 * 1.0 molal = 2.6 molal

3. Calculate the mole fraction of the solute (x_solute):

   For a dilute solution, mole fraction of solute ≈ (molality of solute * molar mass of water) / 1000
   Molar mass of water (H2O) = 18 g/mol
   x_solute = (2.6 * 18) / 1000 = 0.0468

4.  Calculate the mole fraction of the solvent (x_solvent):

    x_solvent = 1 - x_solute
    x_solvent = 1 - 0.0468 = 0.9532

5.  Calculate the vapour pressure of the solution (P_solution):

   P_solution = x_solvent * P°_water, where P°_water is the vapour pressure of pure water.
   P_solution = 0.9532 * 50 mm Hg = 47.66 mm Hg

6.  Round to the nearest integer:

   P_solution ≈ 48 mm Hg

<b>Therefore, the vapour pressure of the 1.0 molal aqueous solution of MgCl2 at 38°C is approximately 48 mm Hg.</b>
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