Question

How much of a change in freezing point would the ocean experience if the glaciers and icebergs melted? Here is what i have so far:

If all of the glaciers and ice caps on earth melted, 24,064,000 cubic kilometers of fresh water would become mixed in with the ocean’s 1,338,000,000 cubic kilometers of saline water.

24,064,000 km^3 + 1,338,000,000 km^3 = 1,362,064,000 km^3

Currently, the ocean’s salinity is 3.5% (http://water.usgs.gov/edu/whyoceansalty.html). The ocean’s salinity would decrease to 3.4% if all of the glaciers and ice caps were to melt.

1,338,000,000 km^3 saline water x .035 = 46,830,000 km^3 salt

46,830,000 km^3 salt / 1,362,064,000 cm^3 = .034 0.034 x 100 = 3.4%

Currently the freezing point of the ocean’s surface is -1.8 degrees Celsius (28.8 degrees Fahrenheit) (http://nsidc.org/cryosphere/seaice/index.html).

I know it's a freezing point depression question and i use

Delta Tf = m x Kf

Kf of water is 1.86 C/m

But I can't figure out how to get 46,830,000 km^3 salt into molality (moles/kg solvent)

Thanks!!

Answer 1

Volume of fresh water = 24,064,000 cubic kilometers

volume of saline water = 1,338,000,000 cubic kilometers

Total volume = 1,362,064,000 km^3

We may assume that density of sea water does not change on addition of fresh water

From sources the density of sea water = 1029 kg/m3 = 1029 X 10^3 Kg / Km^3

Now mass of water before = 1,338,000,000 X 1029 X 10^3 Kg

Mass of water after = 1,362,064,000 X 1029 X 10^3 Kg

Mass of solute before = 0.035 X 1,338,000,000 X 1029 X 10^3

Mass of solute after = 0.034X 1,362,064,000 X 1029 X 10^3

We can calculate the molecular weight of solute as

Depression in freezing point = Kf X mass of solute / molecular weight of solute X mass of solvent

1.8 = 1.86 X mass of solute / molecular weight of solute X mass of solvent

Molecular weight of solute = 1.86 X 0.035 X 1,338,000,000 X 1029 X 10^3 / 1.8 X 1,338,000,000 X 1029 X 10^3 Kg

Now new depression in freezing point let be Tf

Tf = 1.86 X mass of solute / Molecular weight of solute X mass of solvent

Tf = 1.86 X 0.034X 1,362,064,000 X 1029 X 10^3X 1.8 X 1,338,000,000 X 1029 X 10^3 Kg / 1.86 X 0.035 X 1,338,000,000 X 1029 X 10^3 X 1,362,064,000 X 1029 X 10^3 Kg

Tf = 0.034X 1.8 / 0.035 = 1.75

So new freezing point = -1.75 ⁰C