How many grams of CdCl2 * 2H20 crystalline hydrate should be taken to prepare 120 g of a 2.29%

How many grams of CdCl2 * 2H20 crystalline hydrate should be taken to prepare 120 g of a 2.29% aqueous solution of cadmium chloride? Determine the molar concentration of this solution if its density is 1.02 g / ml.

Given:
m solution (CdCl2) = 120 g
ω (CdCl2) = 2.29%
ρ solution (CdCl2) = 1.02 g / ml

To find:
m (CdCl2 * 2H20) -?
See (CdCl2) -?

Decision:
1) Find the mass of the substance CdCl2:
m (CdCl2) = ω (CdCl2) * m solution (CdCl2) / 100% = 2.29% * 120/100% = 2.748 g;

2) Find the amount of substance CdCl2:
n (CdCl2) = m (CdCl2) / Mr (CdCl2) = 2.748 / 183 = 0.015 mol;

3) Find the amount of substance CdCl2 * 2H20:
n (CdCl2 * 2H20) = n (CdCl2) = 0.015 mol;

4) Find the mass of the substance CdCl2 * 2H20:
m (CdCl2 * 2H20) = n (CdCl2 * 2H20) * Mr (CdCl2 * 2H20) = 0.015 * 219 = 3.29 g;

5) Find the volume of the CdCl2 solution:
V solution (CdCl2) = m solution (CdCl2) / ρ solution (CdCl2) = 120 / 1.02 = 117.647 ml =;

6) Find the molar concentration of the CdCl2 solution:
Cm (CdCl2) = n (CdCl2) / V solution (CdCl2) = 0.015 / 0.118 = 0.12 mol / l.

Answer: The mass of CdCl2 * 2H20 is 3.29 g; the molar concentration of the CdCl2 solution is 0.12 mol / l.