Respuesta :
Answer: The molarity of chloride ions in the solution is 0.78 M
Explanation:
- For NaCl:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of NaCl = 1.70 g
Molar mass of NaCl = 58.5 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of NaCl}=\frac{1.70g}{58.5g/mol}=0.029mol[/tex]
1 mole of NaCl produces 1 mole of sodium ions and 1 mole of chloride ions.
Moles of chloride ions = 0.029 moles
- For calcium chloride:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .....(1)
Molarity of calcium chloride solution = 0.100 M
Volume of solution = 50.0 mL
Putting values in equation 1, we get:
[tex]0.100M=\frac{\text{Moles of }CaCl_2\times 1000}{50}\\\\\text{Moles of }CaCl_2=\frac{0.100\times 50}{1000}=0.005mol[/tex]
1 mole of calcium chloride produces 1 mole of calcium ions and 2 moles of chloride ions.
Moles of chloride ions = (2 × 0.005) = 0.01 moles
Now, calculating the molarity of chloride ions in final solution, we use equation 1:
Total moles of chloride ions in solution = (0.029 + 0.01) = 0.039 moles
Total volume of solution = 50.0 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of chloride ions in solution}=\frac{0.039\times 1000}{50.0}\\\\\text{Molarity of solution}=0.78M[/tex]
Hence, the molarity of chloride ions in the solution is 0.78 M
