Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

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


1.92 M

Work Step by Step

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
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