Question

Calculate the volume needed to make 2.20 x 10 2 mL of 0.100 M CaCl 2 solution. What would be the total concentration of the final [Cl - ] alone ?

Answer 1

Given, volume of the solution = 2.2 x 10² mL = 220 mL = 0.22 L
       molarity of the solution = 0.1 M = 0.1 mol/L
       Therefore, no. moles of CaCl₂ = molarity x volume in L
                                                  = 0.1 x 0.22 L
                                                  = 0.022 mol
       no. moles = mass/g.mol.wt
       mass of CaCl₂ = no. moles x g.mol.wt
                              = 0.022 mol x 40.08 g/mol
                              = 0.882 g of CaCl₂
                  
Therefore, 0.882 g of CaCl₂ is needed to make 220 mL of a 0.1 M solution.

Hence, we need 0.882 g of CaCl₂ and 220 mL of water to make the desired solution.

TO find the concentration of [Cl^-] ion alone, we need to look the formula. One CaCl² molecule contains 2 Cl^- ions.
Therefore, the concetration of [Cl^-] ion = 2 x [CaCl₂]
                                       = 2 x 0.1 M
                                       = 0.2 M
                                      
   hence, the concentration of [Cl-] = 0.2 M