Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Concentrations of Solutions - Molarity - Page 110: 63

1.92 M

Molarity = Number of moles of solute/Volume of solution in litres. No. of moles of solute = Mass of NaCl/Molar mass of NaCl = $\frac{4.49g}{58.5gmol^{-1}}$ Volume of solution in litres = $40.0\times10^{-3}L$ Molarity = $\frac{\frac{4.49g}{58.5gmol^{-1}}}{40.0\times10^{-3}L}$ = 1.92 mol/L = 1.92 M