Answer:
There is used 201,655 grams CaCl2 and 448,845 grams of water (H2O)
Explanation:
w% = m(CaCl2) / m(total) x 100%
-> (m(Cacl2) = m(total) x w% ) / 100 %
-> m(CaCl2) = (650,5g x 31%) / 100% = 201,655g
Total mass : m(total) = m(CaCl2) + m(H2O)
-> m(H2O) = m(total) - m(CaCl2)
->m(H2O) = 650,5g - 201,655g = 448,845g