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

26 g sucrose.

Molarity M= 0.15 M Volume of solution in litre V= 0.50 L Number of moles of solute n=MV=$0.15M\times0.50L= 0.075\, mol$ Mass in grams= n$\times$ Molar mass of $C_{12}H_{22}O_{11}$ $= 0.075 mol\times342.3 g/mol=26\, g$