Answer
$28\%\,NH_{3}$
Work Step by Step
14.8 M implies that 14.8 moles of ammonia are present in a 1 L solution.
Mass of solute=$ 14.8\,mol\times\frac{17.031\,g}{1\,mol}=252\,g$
Mass of the solution= Density$\times$Volume
$=\frac{0.90\,g}{1\,cm^{3}}\times\frac{1000\,cm^{3}}{1\,L}\times1\,L=900\,g$
Weight percent=$ \frac{\text{mass of solute}}{\text{mass of solution}}\times100\%$
$=\frac{252\,g}{900\,g}\times100\%=28\%$
