How to use molality and molarity to determine the molar mass of a compound?

Question

An aqueous solution of a certain organic compound has a density of $1.063~\mathrm{g/mL}$, an osmotic pressure of $12.16~\mathrm{atm}$ at $25.0~\mathrm{^\circ{}C}$, and a freezing point of $-1.03~\mathrm{^\circ{}C}$. The compound is known not to dissociate in water. What is the molar mass of the compound?

The molarity and molality can be determined with the given information.

$$m=\dfrac{\Delta T_{f}}{k_{f}}=0.554m$$ and
$$ M=\dfrac{II}{RT}=0.497M$$

$$\text{mass of }\ce{H2O}=\dfrac{1000~\mathrm{g}~\ce{H2O}}{0.554~\mathrm{mol}~\text{of solute}}\cdot 0.497~\mathrm{mol}$$

I don't understand how the last equation is used to determine the mass of $\ce{H2O}$. How are we justified in using the molarity and molality to determine the mass of water?

Answer 1

You can use 3 different approaches to solve that. One I've given in the comments, another one that I will use here that hopefully sheds some light to the equation which by itself out of context here doesn't make much sense.

1. In say 1000g of water you have 0.554 mol of compound (definition of molality)
2. In 1000 ml of solution you have 0.497 mol of compound (definition of molarity)
3. Thus in 1115 ml of solution you have 0.554 mol (linearity)
4. So you have 1000g of water and 0.554 mol of compound in 1115 ml of solution or 1115*1.063 = 1185 g of solution (using the density)
5. Obvioulsy enough you have 1185-1000 = 185g of compound. But we know already that there are 0.554 mol of the compound. So its molecular weight is 185/0.554 = 333.8 (so it's not sugar?)

Same result gives my (preferred) method in the comments

Answer 2

As: $$m=molality\;=\;\frac{moles\; of\; solute}{mass\;of\;solvent}=\frac n{M_B}$$ $$M_B=\frac nm$$ And as : $$M=molarity\;=\;\frac{moles\; of\; solute}{volume\;of\;solution}=\frac n{V}$$ $$n=MV$$ Now: $$M_B=\frac {MV}m$$