What are the molarity AND molality of an ammoniacal solution composed of a mass of 30.6*g with respect to ammonia, and 81.3*g with respect to water, to give a solution whose density is 0.982*g*mL^-1?

Apr 20, 2017

$\text{Molarity"="Moles of solute"/"Volume of solution}$.........

Explanation:

And thus..........................

$\text{Molarity} = \frac{\frac{30.6 \cdot g}{17.01 \cdot g \cdot m o {l}^{-} 1}}{\frac{30.6 + 81.3 \cdot g}{0.982 \cdot g \cdot m {L}^{-} 1}} = \frac{1.80 \cdot m o l}{0.114 \cdot L} = 15.8 \cdot m o l \cdot {L}^{-} 1$.

And $\text{Molality"="Moles of solute"/"Kilograms of solvent}$

$= \frac{\frac{30.6 \cdot g}{17.01 \cdot g \cdot m o {l}^{-} 1}}{81.3 \times {10}^{-} 3 \cdot k g}$ $= \frac{1.80 \cdot m o l}{0.114 \cdot L} = 22.1 \cdot m o l \cdot k {g}^{-} 1$.

I am not overly happy with this question, because an ammoniacal solution of this concentration is almost unreasonable; open this in a lab and the $\text{pen and ink}$ would clear the lab pdq.