What is the molarity of the following?

23g of a solution Ba Br2 in a 2.3L of solution

15g of CaS in 2 L of solution

Molarity = number mole of substances / Volume solution (L)

To find number of mole we need to find Molar mass, at first.

1) M (BaBr2) = M (Ba) + 2 M (Br) = 137.3 + 2*80.0 = 197.3 g/mol

23g BaBr2 * (1mol BaBr2/197.3 g BaBr2) = 0.12 mol BaBr2

Molarity (BaBr2 solution) = 0.12 mol/2.3 L≈0.052 mol/L (BaBr2 solution)

Molarity of BaBr2 solution = 0.052 M

2) M (CaS) = M (Ca) + M (S) = 40.1+32.1=72.2 g/mol

15 g CaS * (1mol CaS/72.2 g CaS) = 0.21 mol CaS

Molarity (CaS solution) = 0.21 mol/2 L≈ 0.11 mol/L (CaS solution)

Molarity of CaS solution = 0.11 M