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Answer :
To find the concentration of [tex]\( Cl^- \)[/tex] ions when two solutions are mixed, follow these steps:
1. Identify Volumes and Concentrations:
- The volume of LiCl solution is 837 mL, with a concentration of 0.463 M.
- The volume of MgCl[tex]\(_2\)[/tex] solution is 913 mL, with a concentration of 0.705 M.
2. Convert Volumes from mL to L:
- Volume of LiCl = 837 mL = 0.837 L
- Volume of MgCl[tex]\(_2\)[/tex] = 913 mL = 0.913 L
3. Calculate Moles of [tex]\( Cl^- \)[/tex] from Each Compound:
- LiCl:
- LiCl dissociates in water to provide 1 mole of [tex]\( Cl^- \)[/tex] per mole of LiCl.
- Moles of [tex]\( Cl^- \)[/tex] from LiCl = concentration of LiCl × volume of LiCl
- Moles of [tex]\( Cl^- \)[/tex] from LiCl = 0.463 M × 0.837 L
- Moles of [tex]\( Cl^- \)[/tex] from LiCl = 0.387531 moles
- MgCl[tex]\(_2\)[/tex]:
- MgCl[tex]\(_2\)[/tex] dissociates to provide 2 moles of [tex]\( Cl^- \)[/tex] per mole of MgCl[tex]\(_2\)[/tex].
- Moles of [tex]\( Cl^- \)[/tex] from MgCl[tex]\(_2\)[/tex] = concentration of MgCl[tex]\(_2\)[/tex] × volume of MgCl[tex]\(_2\)[/tex] × 2
- Moles of [tex]\( Cl^- \)[/tex] from MgCl[tex]\(_2\)[/tex] = 0.705 M × 0.913 L × 2
- Moles of [tex]\( Cl^- \)[/tex] from MgCl[tex]\(_2\)[/tex] = 1.28733 moles
4. Calculate Total Moles of [tex]\( Cl^- \)[/tex]:
- Add the moles of [tex]\( Cl^- \)[/tex] from both LiCl and MgCl[tex]\(_2\)[/tex]:
- Total moles of [tex]\( Cl^- \)[/tex] = 0.387531 moles + 1.28733 moles
- Total moles of [tex]\( Cl^- \)[/tex] = 1.674861 moles
5. Calculate the Total Volume of the Mixed Solution:
- Total volume = 0.837 L (from LiCl) + 0.913 L (from MgCl[tex]\(_2\)[/tex])
- Total volume = 1.75 L
6. Calculate the Concentration of [tex]\( Cl^- \)[/tex]:
- Concentration of [tex]\( Cl^- \)[/tex] = total moles of [tex]\( Cl^- \)[/tex] / total volume
- Concentration of [tex]\( Cl^- \)[/tex] = 1.674861 moles / 1.75 L
- Concentration of [tex]\( Cl^- \)[/tex] = 0.957063 M
Thus, the concentration of [tex]\( Cl^- \)[/tex] ions in the resulting solution is approximately 0.957 M.
1. Identify Volumes and Concentrations:
- The volume of LiCl solution is 837 mL, with a concentration of 0.463 M.
- The volume of MgCl[tex]\(_2\)[/tex] solution is 913 mL, with a concentration of 0.705 M.
2. Convert Volumes from mL to L:
- Volume of LiCl = 837 mL = 0.837 L
- Volume of MgCl[tex]\(_2\)[/tex] = 913 mL = 0.913 L
3. Calculate Moles of [tex]\( Cl^- \)[/tex] from Each Compound:
- LiCl:
- LiCl dissociates in water to provide 1 mole of [tex]\( Cl^- \)[/tex] per mole of LiCl.
- Moles of [tex]\( Cl^- \)[/tex] from LiCl = concentration of LiCl × volume of LiCl
- Moles of [tex]\( Cl^- \)[/tex] from LiCl = 0.463 M × 0.837 L
- Moles of [tex]\( Cl^- \)[/tex] from LiCl = 0.387531 moles
- MgCl[tex]\(_2\)[/tex]:
- MgCl[tex]\(_2\)[/tex] dissociates to provide 2 moles of [tex]\( Cl^- \)[/tex] per mole of MgCl[tex]\(_2\)[/tex].
- Moles of [tex]\( Cl^- \)[/tex] from MgCl[tex]\(_2\)[/tex] = concentration of MgCl[tex]\(_2\)[/tex] × volume of MgCl[tex]\(_2\)[/tex] × 2
- Moles of [tex]\( Cl^- \)[/tex] from MgCl[tex]\(_2\)[/tex] = 0.705 M × 0.913 L × 2
- Moles of [tex]\( Cl^- \)[/tex] from MgCl[tex]\(_2\)[/tex] = 1.28733 moles
4. Calculate Total Moles of [tex]\( Cl^- \)[/tex]:
- Add the moles of [tex]\( Cl^- \)[/tex] from both LiCl and MgCl[tex]\(_2\)[/tex]:
- Total moles of [tex]\( Cl^- \)[/tex] = 0.387531 moles + 1.28733 moles
- Total moles of [tex]\( Cl^- \)[/tex] = 1.674861 moles
5. Calculate the Total Volume of the Mixed Solution:
- Total volume = 0.837 L (from LiCl) + 0.913 L (from MgCl[tex]\(_2\)[/tex])
- Total volume = 1.75 L
6. Calculate the Concentration of [tex]\( Cl^- \)[/tex]:
- Concentration of [tex]\( Cl^- \)[/tex] = total moles of [tex]\( Cl^- \)[/tex] / total volume
- Concentration of [tex]\( Cl^- \)[/tex] = 1.674861 moles / 1.75 L
- Concentration of [tex]\( Cl^- \)[/tex] = 0.957063 M
Thus, the concentration of [tex]\( Cl^- \)[/tex] ions in the resulting solution is approximately 0.957 M.
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