what would the molarity of a solution of 125g of NaHCO3 in enough water to make 675ml of solution?

Molar mass of NaHCO₃ = (23.0 + 1.0 + 12.0 + 16.0×3) g/mol = 84.0 g/mol
No. of moles of NaHCO₃ = (125 g) / (84.0 g/mol) = 1.488 mol

Volume of the solution = 675 ml = 0.675 L

Molarity of the solution = (1.488 mol) / (0.675 L) = 2.20 M