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

Explanation :

First we have to calculate the mole of and .

Molar mass of NaCl = 58.9 g/mol

and,

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

So, 0.005 mole of 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.

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

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)