# What is the "molality" of a 0.030*g mass of sugar dissolved in 300*mL of water?

May 3, 2017

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

#### Explanation:

And so "Molality"=((0.03*g)/(180.16*g*mol^-1))/(0.3*kg)=??mol*kg^-1.

At these concentrations the $\text{molality}$ would be identical to the $\text{molarity}$.

Note that $1 \cdot {\mathrm{dm}}^{3}$ is a fancy way of saying $1 \cdot L$, $1 \cdot {\mathrm{dm}}^{3}$ $=$ $\text{1 decimetre}$, where the prefix $\text{deci} \equiv {10}^{-} 1$.

And thus 1*dm^3"-=(10^-1*m)^3=10^-3*m^3=1*L, because there are $1000 \cdot L \cdot {m}^{-} 3$. A $\text{cubic metre}$ is a huge volume.