Calculate the freezing point of a solution of 500.0 g of ethylene glycol dissovled in 500g water. Kf = 1.86 degrees C/m and Kb (which my instructor said was just extraneous info that is not used here) is 0.512 degrees C/m.

= -29.96 °C

Explanation: A solution has a lower freezing point compared to the pure solvent, due to the solute lowering the vapor pressure of the solvent.

The "new" freezing point is calculated as:

is the molal freezing-point depression constant

m is molality concentration, i.e., moles of solute per kilogram of solvent

Before calculating freezing point, let's find moles of ethylene glycol () in the solution:

Molar mass = 62.08 g/mol

For 500 g:

n = 8.05 moles

The molality concentration for 0.5kg of water:

m =

m = 16.11

The freezing point will be:

Freezing point of a solution of Ethylene Glycol and Water is -29.96°C

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

gold is my favorite solid.