Question

How much gram of urea dissolved in 100 gm of water to create a difference of 105°C between freezing point and boiling point?

Answer 1

Consider urea,

Recall,

`DeltaT_"b" = iK_"b"m`

We need the boiling point to be `0+105°"C" = 105°"C"`, thus, `DeltaT_"b" = 5°"C"`

`i = 1`, urea won't measurably dissociate in water, and

`K_"b" = (0.512°"C")/m`

Hence, we need,

`DeltaT_"b" = iK_"b"n/m_(H_2O)`

`=> n = (DeltaT_"b" * m_(H_2O))/(iK_"b") approx 0.977"mol" * (60.06"g")/"mol" approx 58.7"g"`

of urea dissolved in that volume of water to raise the boiling point by `5°"C"`.

Answer 2

You must use 13 g of urea.

Explanation:

For pure water, the difference between the boiling point and freezing point
is 100.00 °C.

The added urea will simultaneously raise the boiling point and lower the freezing point, so

`"Boiling point elevation + freezing point depression = 5 °C"`

The formula for boiling point elevation `ΔT_"b"` is

`color(blue)(bar(ul(|color(white)(a/a)ΔT_"b" = iK_"b"bcolor(white)(a/a)|)))" "`

where

• `i` is the van't Hoff `i` factor
• `K_"b"` is the molal boiling point elevation constant
• `b` is the molal concentration of the solution

The formula for freezing point depression `ΔT_"f"` is

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

where

• `K_"b"` is the molal boiling point elevation constant

Then,

`ΔT_text(b) + ΔT_text(f) = iK_text(b)b + iK_text(f)b = "5 °C"`

`i(K_text(b) + K_text(f))b = "5 °C"`

In this problem,

`i color(white)(m)= 1`, because urea is a nonelectrolyte
`K_text(b) = "0.515 °C·kg·mol"^"-1"`
`K_text(f)color(white)(l) = "1.853 °C kg·mol"^"-1"`

`1("0.515 °C·kg·mol"^"-1" + "1.853 °C·kg·mol"^"-1")b`

`= 2.368b color(red)(cancel(color(black)("°C")))"·kg·mol"^"-1" = 5 color(red)(cancel(color(black)("°C")))`

`b = 5/("2.368 kg·mol"^"-1") = "2.1 mol/kg"`

`"Moles of urea" = 0.100 color(red)(cancel(color(black)("kg"))) × "2.1 mol"/(1 color(red)(cancel(color(black)("kg")))) = "0.21 mol"`

`"Mass of urea" = 0.21 color(red)(cancel(color(black)("mol urea"))) × "60.06 g urea"/(1 color(red)(cancel(color(black)("mol urea")))) = "13 g urea"`

Check:

`ΔT_text(b) = "0.515 °C"·color(red)(cancel(color(black)("kg·mol"^"-1"))) × 2.1 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "1.1 °C"`

`ΔT_text(f) = "1.853 °C"·color(red)(cancel(color(black)("kg·mol"^"-1"))) × 2.1 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "3.9 °C"`

`ΔT_text(b) + ΔT_text(f) = "1.1 °C + 3.9 °C = 5 °C"`

It checks!