Answer
160. g $NH_{4}Cl$
Work Step by Step
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 $NH_{4}Cl$
$= 3.00 mol\times53.491 g/mol= 160. g$