What is the "molality" of a solution composed of 684*g of sucrose dissolved in a mass of 1000*g of water?

$\text{Molality"="Moles of solute"/"Kilograms of solvent..}$
$\frac{\frac{684 \cdot g}{342.3 \cdot g \cdot m o {l}^{-} 1}}{1.000 \cdot k g} = 2.00 \cdot m o l \cdot k {g}^{-} 1$.
We would need the density of the solution to convert this to the $\text{molarity}$, which is specified by the quotient, $\text{molarity"="moles of solute"/"volume of solution}$. The values would not be too much different.