Answer
$15\;M\;NH_3$
Work Step by Step
28% by mass means that there are 28 g $NH_3$ (molar mass = 17.03 g/mol) per 100 g solution.
Molarity = moles solute / L solution
First, calculate moles solute:
$\frac{28\;g\;NH_3}{}(\frac{1\;mol}{17.03\;g})=1.644\;mol\;NH_3$
Next, calculate volume of solution in L:
$\frac{100\;g\;solution}{}(\frac{1\;mL}{0.90\;g})(\frac{1\;L}{1000 mL})=0.1111\;L\;solution$
Next, calculate molarity.
$\frac{1.644\;mol\;NH_3}{0.1111\;L\;solution}=15\;M\;NH_3$
