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

160. g $NH_{4}Cl$

Molarity M= 4.00 M Volume of solution in litre= 0.750 L Number of moles of solute n=MV=$4.00M\times.750L= 3.00 mol$ Mass in grams= n$\times$ Molar mass of $NH_{4}Cl$ $= 3.00 mol\times53.491 g/mol= 160. g$