Solveeit Logo
Login

Question

Question: 3 moles of a sparingly soluble salt $PbCl_2$ dissociates in 1000 mL solution to give 2 mole of chlor...

3 moles of a sparingly soluble salt PbCl2PbCl_2 dissociates in 1000 mL solution to give 2 mole of chloride ions. What is the osmotic pressure of solution at 27°C?

A

73.89 atm

B

60.42 atm

C

96.52 atm

D

50.50 atm

Answer

73.89 atm

Explanation

Solution

The sparingly soluble salt is PbCl2PbCl_2. When it dissolves in water, it dissociates into ions according to the equation:

PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)

The problem states that 2 moles of chloride ions (Cl−Cl^-) are produced in the 1000 mL solution. According to the dissociation equation, for every 2 moles of Cl−Cl^- ions produced, 1 mole of Pb2+Pb^{2+} ions is produced, and 1 mole of PbCl2PbCl_2 must have dissolved.

So, the number of moles of Pb2+Pb^{2+} ions in the solution is: Moles of Pb2+=1 mole Pb2+2 moles Cl−×2 moles Cl−=1 molePb^{2+} = \frac{1 \text{ mole } Pb^{2+}}{2 \text{ moles } Cl^-} \times 2 \text{ moles } Cl^- = 1 \text{ mole}.

The total number of moles of ions in the solution is the sum of the moles of Pb2+Pb^{2+} ions and Cl−Cl^- ions: Total moles of ions = Moles of Pb2+Pb^{2+} + Moles of Cl−Cl^- = 1 mole + 2 moles = 3 moles.

The volume of the solution is given as 1000 mL, which is equal to 1 L.

The total molar concentration of ions in the solution is: Cions=Total moles of ionsVolume of solution (L)=3 moles1 L=3 MC_{ions} = \frac{\text{Total moles of ions}}{\text{Volume of solution (L)}} = \frac{3 \text{ moles}}{1 \text{ L}} = 3 \text{ M}.

The temperature of the solution is given as 27°C. We need to convert this temperature to Kelvin: T(K)=T(°C)+273.15T(K) = T(°C) + 273.15 T=27+273=300 KT = 27 + 273 = 300 \text{ K}. (Using 273 for simplicity as is common in such problems, leading to the provided options).

The osmotic pressure (π\pi) of a solution is given by the formula: π=CionsRT\pi = C_{ions}RT where:

  • CionsC_{ions} is the total molar concentration of ions.
  • RR is the ideal gas constant, R=0.0821 L atm mol−1 K−1R = 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}.
  • TT is the temperature in Kelvin.

Plugging in the values: π=(3 mol/L)×(0.0821 L atm mol−1 K−1)×(300 K)\pi = (3 \text{ mol/L}) \times (0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K}) π=3×0.0821×300 atm\pi = 3 \times 0.0821 \times 300 \text{ atm} π=900×0.0821 atm\pi = 900 \times 0.0821 \text{ atm} π=73.89 atm\pi = 73.89 \text{ atm}.

The osmotic pressure of the solution is 73.89 atm.