Pratical Dilemmas?

Question

>The FW of anhydrous CaCl2 is 111.0, so FW for CaCl2 dihydrate would be 147.0 (yes or No)
>If you had a protocol that called for 50g. anhydrous CaCl2, how would you make this solution using anhydrous CaCl2.
50mM/1000mL X 111.0 g = 5.55 g of CaCl2 to be dissolved in 1L vessel.
>How would you make this solution using CaCl2 dihydrate ?
50mM/1000mL X 147.0 g = 7.35 g of CaCl2.2H20

I will appreciate your input.

Answer 1

Your calculations are all very good. I think you slipped up, though, in typing in the question. It doesn't read quite right.