[needhelp22]

we are doing applications of aqueous equilibria in our chem class this week and we have a challenge problem for homework

"You add an excess of solid MX in 250 g of water. You measure hte freezing point and find it to be -.028 degrees celsius. What is the Ksp of the sold? Assume the density of the solution is 1.0 g/cm cubed"

i have no idea where to start, i just need a little help to what direction i should be going, thanks.

[Borek]

Colligative properties. Freezing point depression depends on the concentration of dissolved substance.

[needhelp22]

arite im stuck

i used

(change in)T = K(f) times molality

i figured out the molality to be .015 moles/kg

using the density i used stoichiometry to find out the moles of MX

.015 moles of MX|.250 kg of water
----------------|-----------------------------
1 kg of water     |

i got .0038 moles of MX

knowing that 1 kg of water = 1 L of water i found molarity of MX

.0038 moles/.250 L = .0151 Molarity

now finding ksp is the real question

so i was thinking this...


MX --> M + X

ksp = [M][X]

but i only know hte molarites of MX, is it the same for M and for X (since the ratio's 1 to 1)

thanks

[Borek]

M=X is OK. Have you remembered about Van 't Hoff factor?