Question

A solution is prepared by adding 0.700 g of solid NaClNaCl to 50.0 mL of 0.100 M CaCl2CaCl2. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

Answer 1

Answer : The molarity of chloride ion in the final solution is, 0.436 M

Explanation :

First we have to calculate the mole of $NaCl$ and $CaCl_2$.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}$

Molar mass of NaCl = 58.9 g/mol

$\text{Moles of }NaCl=\frac{0.700g}{58.9g/mol}=0.0118mol$

and,

$\text{Moles of }CaCl_2=\text{Concentration of }CaCl_2\times \text{Volume of solution}$

$\text{Moles of }CaCl_2=0.100M\times 0.050L=0.005mol$

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 $CaCl_2$ dissociates to give 1 mole of sodium ion and 2 mole of chloride ion.

So, 0.005 mole of $CaCl_2$ 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.

$\text{Molarity}=\frac{\text{Total moles}}{\text{Total volume}}$

$\text{Molarity}=\frac{0.0218mol}{0.050L}=0.436M$

Thus, the molarity of chloride ion in the final solution is, 0.436 M