2) What is the molality of solution that contains 50 g ethanol (C,H,OH) in 325 g water? The molality of a solution is equal to the moles of solute divided by the mass of solvent in kilograms. The answer is 3.338 mol/kg| 3) What will be the freezing point change for the solution in question 2?

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{enthanol}=50\mathrm{g}: \mathrm{mass} \mathrm{of} \mathrm{solvent}=325\mathrm{g} \\ \mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{ethanol}=46.07\mathrm{g}/\mathrm{mol} \mathrm{i}.\mathrm{e}, 46.07\mathrm{g}/\mathrm{mol} \\ \mathrm{no}. \mathrm{of} \mathrm{moles} \mathrm{ethanol}=\frac{50}{46.07} \\ \mathrm{Molarity}, \mathrm{m}=\frac{\mathrm{no}.\mathrm{of} \mathrm{moles}}{\mathrm{mass} \mathrm{of} \mathrm{solvent}\left(\mathrm{g}\right)}\times 1000 \\ =\frac{50}{46.07\times 325}\times 1000 \mathrm{i}.\mathrm{e}, \frac{50\times 1000}{46.07\times \times 325} \\ =3.339\mathrm{g}/\mathrm{mol} \mathrm{i}.\mathrm{e}, 3.339 \mathrm{g}/\mathrm{mol} \\ ∆{\mathrm{T}}_{\mathrm{F}}={\mathrm{K}}_{\mathrm{F}} \mathrm{m} {\mathrm{K}}_{\mathrm{F}} \mathrm{for} \mathrm{water}=1.86 \\ ∆{\mathrm{T}}_{\mathrm{F}}=1.86\times 3.339 \\ ∆{\mathrm{T}}_{\mathrm{F}}=6.21 \\ ∆{\mathrm{T}}_{\mathrm{F}}=\mathrm{T}{°}_{\mathrm{F}}-{\mathrm{T}}_{\mathrm{F}} \mathrm{where} \mathrm{T}{°}_{\mathrm{F}}=\mathrm{freezing} \mathrm{point} \mathrm{of} {\mathrm{H}}_2\mathrm{O}\left(\mathrm{pure}\right) \\ 6.21=0-{\mathrm{T}}_{\mathrm{F}} {\mathrm{T}}_{\mathrm{F}}=\mathrm{freezing} \mathrm{point} \mathrm{of} \mathrm{solution}. \\ {\mathrm{T}}_{\mathrm{F}}=-6.21°\mathrm{C} \\ \mathrm{freezing} \mathrm{point} \mathrm{change} \mathrm{by} 6.21°\mathrm{C} \mathrm{for} \mathrm{solution}. \end{gathered}$$