Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605c: 44c

7.46 g $Na_{2}SO_{4}$

Molarity M= 0.350 M Volume of solution in litre V= 0.150 L Number of moles of solute n=MV=$0.350M\times.150L= 0.0525 mol$ Mass in grams= n$\times$ Molar mass of $Na_{2}SO_{4}$ $= 0.0525 mol\times142.04 g/mol=7.46 g$