Chemistry 10th Edition

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