Respuesta :
Answer : The molarity of chloride ion in the final solution is, 0.436 M
Explanation :
First we have to calculate the mole of [tex]NaCl[/tex] and [tex]CaCl_2[/tex].
[tex]\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}[/tex]
Molar mass of NaCl = 58.9 g/mol
[tex]\text{Moles of }NaCl=\frac{0.700g}{58.9g/mol}=0.0118mol[/tex]
and,
[tex]\text{Moles of }CaCl_2=\text{Concentration of }CaCl_2\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }CaCl_2=0.100M\times 0.050L=0.005mol[/tex]
Now we have to calculate the moles of chloride ion.
As, 1 mole of NaCl dissociates to give 1 mole of sodium ion and 1 mole of chloride ion.
So, 0.0118 mole of NaCl dissociates to give 0.0118 mole of sodium ion and 0.0118 mole of chloride ion.
and,
As, 1 mole of [tex]CaCl_2[/tex] dissociates to give 1 mole of sodium ion and 2 mole of chloride ion.
So, 0.005 mole of [tex]CaCl_2[/tex] dissociates to give 0.005 mole of sodium ion and (2×0.005=0.01) mole of chloride ion.
Now we have to calculate the total moles of chloride ion and volume of solution.
Total moles of chloride ion = 0.0118 + 0.01 = 0.0218 mol
Total volume of solution = 50.0 mL = 0.050 L
Now we have to calculate the molarity of chloride ion in the final solution.
[tex]\text{Molarity}=\frac{\text{Total moles}}{\text{Total volume}}[/tex]
[tex]\text{Molarity}=\frac{0.0218mol}{0.050L}=0.436M[/tex]
Thus, the molarity of chloride ion in the final solution is, 0.436 M
