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

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?

• Try dividing the unit m by the unit M and see what you get. – Brinn Belyea Aug 2 '14 at 20:32
• It's impossible to find the mass of water. The problem is asking for the molar mass of the compound. Assume 1 mol of compound, use the definitions of molarity and molality, find the weight of the compound and as you assumed 1 mol of it, the number you found is going to be the molecular mass. – K_P Aug 3 '14 at 12:44

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: $$M_B=\frac {MV}m$$