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: 62

Answer

0.866 M

Work Step by Step

Mass of $Na_{3}PO_{4}$ = 355 g Molar mass of $Na_{3}PO_{4}$ = 163.94 g/mol Volume in litres = 2.50 L Molarity = $\frac{Mass\div Molar.mass}{Volume(L)} = \frac{355g\div163.94gmol^{-1}}{2.50 L}$ = 0.866 mol/L = 0.866 M
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