What volume of ethanol (density = 0.7893 g/cm^{3}) should be added to 450 mL of water in order to have a solution that freezes at -15.0°C? [For water, K_{f} = 1.86°C/m.]

A) 371 mL

B) 470 mL

C) 212 mL

D) 132 mL

E) 167 mL

Solution: What volume of ethanol (density = 0.7893 g/cm3) should be added to 450 mL of water in order to have a solution that freezes at -15.0°C? [For water, Kf = 1.86°C/m.]A) 371 mLB) 470 mLC) 212 mLD) 132 mLE

What volume of ethanol (density = 0.7893 g/cm^{3}) should be added to 450 mL of water in order to have a solution that freezes at -15.0°C? [For water, K_{f} = 1.86°C/m.]

A) 371 mL

B) 470 mL

C) 212 mL

D) 132 mL

E) 167 mL

We’re being asked to **determine the volume of ethanol (C _{2}H_{5}OH)** that must be added to

Recall that the freezing point of a solution is *lower* than that of the pure solvent and the ** change in freezing point (ΔT_{f})** is given by:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{pure}\mathbf{}\mathbf{solvent}}{\mathbf{-}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{solution}}}$

The ** change in freezing point** is also related to the molality of the solution:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{imK}}}_{{\mathbf{f}}}}$

where:

**i** = van’t Hoff factor

**m** = molality of the solution (in m or mol/kg)

**K _{f}** = freezing point depression constant (in ˚C/m)

Recall that the ** molality of a solution** is given by:

$\overline{){\mathbf{molality}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}}$

**For this problem, we need to do the following:**

* Step 1:* Calculate for ΔT

* Step 2:* Determine the molality of the solution.

* Step 3:* Calculate the volume of C