Chemistry: The Central Science (13th Edition)

$15\;M\;NH_3$
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. $\frac{1.644\;mol\;NH_3}{0.1111\;L\;solution}=15\;M\;NH_3$