1.SOLUBILITY:
At 25C, the molar solubility of calcium phosphate in water is 1.1*10^{-7}M. Calculate the solubility in grams per liter.
I m not sure how to convert M to g/L
MOLAR MASS OF COLLIGATIVE PROPERTIES:
2.tert-Butyl alcohol is a solvent with a Kf of 9.10 C/m and a freezing point of 25.5 C. When 0.807g of an unknown colorless liquid was dissolved in 11.6 g of tert-butyl alcohol, the solution froze at 15.3 C.
Which of the following is most likely the identity of this unknown liquid?
a.ethylene glycol(molar mass=62.07g/mol)
b.1-octanol (molar mass = 130.22 g/mol)
c.glycerol (molar mass = 92.09 g/mol)
d.2-pentanone (molar mass = 86.13 g/mol)
e.1-butanol (molar mass = 74.12 g/mol)
so, molar mass= mass in grams/# of moles
I think im suppose to find the number of moles, using the given information :
Do i use Pi= MRT, and im not sure what Kf stands for.
4 answers
for the 1st question
solubility =3.41×10−5 g/l
mols = g/molar mass; therefore,
grams = mols*molar mass.
The answer for the solubility copmes out in units of molarity and that is mols/L. Therefore, all you need to do is to convert mols to grams and that will be g/L since M is mols/L.