# What is the molality of a solution that contains 63.0 g HNC_3 in 0.500 kg H_2O?

What do you mean by $H N {C}_{3}$, do you mean ${H}_{2} N C {H}_{3}$, i.e. $\text{methylamine}$?
If you mean $\text{methylamine}$, then the molality of the solution is given by:
$\text{Moles of solute"/"Kilograms of solvent} = \frac{\frac{63.0 \cdot g}{31.06 \cdot g \cdot m o {l}^{-} 1}}{0.500 \cdot k g} = 4.06 \cdot m o l \cdot k {g}^{-} 1$.