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

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