Hello Candidate,

The Total mass of MgCl2 is equal to 95 amu.

So, as the salt contains the degree of +2 charge, hence the equivalent weight of MgCl2 is equal to 50 amu.

The equivalent weight of CaCo3 is equal to 50 amu.

So, 50 mg of MgCl2= 50 mg of CaCo3. The total amount of CaCo3 equals 50× 10^-3 g/L.

Hence, the hardness of CaCo3 in ppm= 5×10^-2/1000 × 10^6= 5 ppm.

Hope that this answer helps you!!