Answer: 0.4533mol/L
Explanation:
Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol
68g of CaCO3 dissolves in 1.5L of solution.
Xg of CaCO3 will dissolve in 1L i.e
Xg of CaCO3 = 68/1.5 = 45.33g/L
Molarity = Mass conc.(g/L) / molar Mass
Molarity = 45.33/100 = 0.4533mol/L
