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Question: A solution containing 200 ml 0.5M KCl is mixed with 50 ml 19% w/v $MgCl_2$ and resulting is diluted ...

A solution containing 200 ml 0.5M KCl is mixed with 50 ml 19% w/v MgCl2MgCl_2 and resulting is diluted 8 times. Molarity of chloride ion is final solution is :

Answer

0.15 M

Explanation

Solution

To find the molarity of chloride ions in the final solution, we need to calculate the total moles of chloride ions from both sources and the final total volume after dilution.

1. Moles of chloride ions from KCl solution: Given:

  • Volume of KCl solution = 200 mL = 0.2 L
  • Molarity of KCl solution = 0.5 M

KCl dissociates as: KCl→K++Cl−\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- From the stoichiometry, 1 mole of KCl produces 1 mole of Cl−\text{Cl}^- ions.

Moles of KCl = Molarity × Volume (L) Moles of KCl = 0.5 mol/L × 0.2 L = 0.1 mol Therefore, moles of Cl−\text{Cl}^- from KCl = 0.1 mol.

2. Moles of chloride ions from MgCl2\text{MgCl}_2 solution: Given:

  • Volume of MgCl2\text{MgCl}_2 solution = 50 mL
  • Concentration of MgCl2\text{MgCl}_2 solution = 19% w/v

19% w/v means 19 grams of MgCl2\text{MgCl}_2 are present in 100 mL of solution. So, the mass of MgCl2\text{MgCl}_2 in 50 mL of solution is: Mass of MgCl2=(19 g100 mL)×50 mL=9.5 g\text{MgCl}_2 = \left(\frac{19 \text{ g}}{100 \text{ mL}}\right) \times 50 \text{ mL} = 9.5 \text{ g}

Now, calculate the moles of MgCl2\text{MgCl}_2. Molar mass of MgCl2\text{MgCl}_2: (Approximate atomic masses: Mg = 24 g/mol, Cl = 35.5 g/mol) Molar mass of MgCl2=24+(2×35.5)=24+71=95 g/mol\text{MgCl}_2 = 24 + (2 \times 35.5) = 24 + 71 = 95 \text{ g/mol}

Moles of MgCl2=MassMolar mass=9.5 g95 g/mol=0.1 mol\text{MgCl}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{9.5 \text{ g}}{95 \text{ g/mol}} = 0.1 \text{ mol}

MgCl2\text{MgCl}_2 dissociates as: MgCl2→Mg2++2Cl−\text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^- From the stoichiometry, 1 mole of MgCl2\text{MgCl}_2 produces 2 moles of Cl−\text{Cl}^- ions.

Moles of Cl−\text{Cl}^- from MgCl2=2×Moles of MgCl2=2×0.1 mol=0.2 mol\text{MgCl}_2 = 2 \times \text{Moles of } \text{MgCl}_2 = 2 \times 0.1 \text{ mol} = 0.2 \text{ mol}.

3. Total moles of chloride ions: Total moles of Cl−=Moles of Cl− from KCl+Moles of Cl− from MgCl2\text{Cl}^- = \text{Moles of } \text{Cl}^- \text{ from KCl} + \text{Moles of } \text{Cl}^- \text{ from } \text{MgCl}_2 Total moles of Cl−=0.1 mol+0.2 mol=0.3 mol\text{Cl}^- = 0.1 \text{ mol} + 0.2 \text{ mol} = 0.3 \text{ mol}.

4. Initial total volume of the mixed solution: Initial total volume = Volume of KCl solution + Volume of MgCl2\text{MgCl}_2 solution Initial total volume = 200 mL + 50 mL = 250 mL.

5. Final volume after dilution: The resulting solution is diluted 8 times. Final volume = Initial total volume × Dilution factor Final volume = 250 \text{ mL} \times 8 = 2000 \text{ mL} = 2 \text{ L}$.

6. Molarity of chloride ion in the final solution: Molarity of Cl−=Total moles of Cl−Final volume (L)\text{Cl}^- = \frac{\text{Total moles of } \text{Cl}^-}{\text{Final volume (L)}} Molarity of Cl−=0.3 mol2 L=0.15 M\text{Cl}^- = \frac{0.3 \text{ mol}}{2 \text{ L}} = 0.15 \text{ M}.

The final molarity of chloride ion in the solution is 0.15 M.

Explanation of the solution:

  1. Calculate moles of Cl−\text{Cl}^- from KCl: 0.5 M×0.2 L=0.1 mol0.5 \text{ M} \times 0.2 \text{ L} = 0.1 \text{ mol}.
  2. Calculate moles of Cl−\text{Cl}^- from MgCl2\text{MgCl}_2: Mass of MgCl2=(19/100)×50 g=9.5 g\text{MgCl}_2 = (19/100) \times 50 \text{ g} = 9.5 \text{ g}. Moles of MgCl2=9.5 g/95 g/mol=0.1 mol\text{MgCl}_2 = 9.5 \text{ g} / 95 \text{ g/mol} = 0.1 \text{ mol}. Since MgCl2\text{MgCl}_2 yields 2 Cl−\text{Cl}^- ions, moles of Cl−=2×0.1 mol=0.2 mol\text{Cl}^- = 2 \times 0.1 \text{ mol} = 0.2 \text{ mol}.
  3. Total moles of Cl−=0.1 mol+0.2 mol=0.3 mol\text{Cl}^- = 0.1 \text{ mol} + 0.2 \text{ mol} = 0.3 \text{ mol}.
  4. Initial volume = 200 mL+50 mL=250 mL200 \text{ mL} + 50 \text{ mL} = 250 \text{ mL}.
  5. Final volume after 8 times dilution = 250 mL×8=2000 mL=2 L250 \text{ mL} \times 8 = 2000 \text{ mL} = 2 \text{ L}.
  6. Final molarity of Cl−=Total moles of Cl−/Final volume=0.3 mol/2 L=0.15 M\text{Cl}^- = \text{Total moles of } \text{Cl}^- / \text{Final volume} = 0.3 \text{ mol} / 2 \text{ L} = 0.15 \text{ M}.