# What is the "molality" of a 10*g mass of salt, whose formula mass was 36.46*g*mol^-1, that is dissolved in a 100*g mass of solvent?

Jul 26, 2017

$\text{Molality} < 3 \cdot m o l \cdot k {g}^{-} 1.$

#### Explanation:

By definition, $\text{Molality"="Moles of solute"/"Kilograms of solvent}$.

Here you have a 10% solution, and here I ASSUME that you mean

w/wxx100%, i.e. "Mass of solute"/"Mass of solvent"xx100%

And so we have a $10 \cdot g$ mass of solute, in a $100 \cdot g$ mass of solvent (which is reasonably assumed to have a $100 \cdot m L$ volume; this is another unspoken assumption; the solvent is clearly water).

And finally $\text{molality} \equiv \frac{\frac{10 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1}}{0.100 \cdot k g} = 2.74 \cdot m o l \cdot k {g}^{-} 1$.

Note that at these concentrations, $\text{molality}$ is equivalent to $\text{molarity}$.