Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605c: 45c

3.4 g $NaHCO_{3}$

Molarity M= 0.20 M Volume of solution in litre V= 0.200 L Number of moles of solute n=MV=$0.20\,M\times.200\,L= 0.040\, mol$ Mass in grams= n$\times$ Molar mass of $NaHCO_{3}$ $= 0.040\,mol\times84.0\, g/mol=3.4\, g$