Question

Question #a0ccf

Answer 1

The student had to use 3.2 L of methanol.

Explanation:

Step 1. Calculate the molality of the solution.

The formula for calculating freezing point depression is

`color(blue)(bar(ul(|color(white)(a/a)ΔT_"f" = K_fmcolor(white)(a/a)|)))" "`

where

`ΔT_"f"` is the change in freezing point

`K_"f"` is the molal freezing point depression constant

`"m"` is the molality of the solution.

We can rearrange the above formula to get

`m = (ΔT_"f")/ K_"f"`

In this problem,

`ΔT_"f" = T_"f"^° -T_"i" = "(-30 °C) -0 °C" = "-30 °C"`

`K_"f" = "-1.86 °C·kg·mol"^"-1"`

∴ `m = ("-30" "°C")/("-1.86" "°C"·"kg·mol"^"-1") = "16.1 mol·kg"^"-1"`

Step 2. Calculate the mass of the solution.

When we mix two liquids, we cannot assume that their volumes are additive,

For example, 1 L of methanol + 2 L of water gives 3.1 L of solution.

This complicates the calculation.

(a) Calculate the percent composition of the solution.

`16.1 color(red)(cancel(color(black)("mol MeOH"))) × "32.04 g MeOH"/(1 color(red)(cancel(color(black)("mol MeOH")))) = "516 g MeOH"`

`"Total mass" = "516 g + 1000 g" = "1516 g"`

`"% MeOH by mass" = "Mass of MeOH"/"Total mass" × 100 % = (516 color(red)(cancel(color(black)("g"))))/(1516 color(red)(cancel(color(black)("g")))) × 100 % = 34.0 %`

(b) Look up the density of the solution.

The density of 34.0 % aqueous MeOH is 0.945 g/mL.

(c) Calculate the mass of the solution.

`"Mass" = 8000 color(red)(cancel(color(black)("mL"))) × "0.945 g"/(1 color(red)(cancel(color(black)("mL")))) = "7560 g"`

Step 3. Calculate the mass of methanol

`"Mass of MeOH" = 7560 color(red)(cancel(color(black)("g soln"))) × "516 g MeOH"/(1516 color(red)(cancel(color(black)("g soln")))) = "2570 g MeOH"`

Step 4. Calculate the volume of methanol

`"Volume of MeOH" = 2570 color(red)(cancel(color(black)("g MeOH"))) × "1 mL MeOH"/(0.792 color(red)(cancel(color(black)("g MeOH")))) = "3200 mL MeOH" = "3.2 L MeOH"`

Answer 2

The student had to use 3.2 L of methanol.

Explanation:

This question appears to assume (incorrectly) that the volumes on mixing of liquids are additive.

We can use only two significant figures, so we might get away with it.

Step 1. Calculate the molality of the solution.

`color(blue)(bar(ul(|color(white)(a/a)ΔT_"f" = K_fmcolor(white)(a/a)|)))" "`

`m = (ΔT_"f")/ K_"f" = ("-30" "°C")/("-1.86" "°C"·"kg·mol"^"-1") = "16.1 mol·kg"^"-1"`

Step 2. Calculate the volumes of methanol and water in the above solution.

`"Volume of MeOH" = 16.1 color(red)(cancel(color(black)("mol MeOH"))) × "1 L MeOH"/(24.7 color(red)(cancel(color(black)("mol MeOH")))) = "0.652 L"`

`"Volume of water" = 1000 color(red)(cancel(color(black)("g water"))) × "1.00 mL water"/(1 color(red)(cancel(color(black)("mL water")))) = "1000 mL water" = "1.00 L water"`

`"Volume of solution" = "0.652 L + 1.00 L" = "1.65 L"`

Step 3. Calculate the volume of methanol in 8.0 L of solution

`"Volume of MeOH" = 8.0 color(red)(cancel(color(black)("L soln"))) × "0.652 L MeOH"/(1.652 color(red)(cancel(color(black)("L soln")))) = "3.2 L MeOH"`

It worked! We get the correct answer with much less work.