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.

for 2nd question answer is . (a)--ehtylene glycol .. just did my mastering chemistry assignmnet stuck on crystals one .. if u can help me out .. plz..

for the 1st question
solubility =3.41×10−5 g/l

the coordination things i don't know how to do that...if i find out ill let you know thanks

molarity = mols/L and

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.

thank you

To convert molar solubility (in M) to solubility (in g/L), you need to know the molar mass of the substance in question. Here's how you can calculate the solubility in this case:

Step 1: Calculate the number of moles of calcium phosphate (Ca3(PO4)2) in the given molar solubility. Molar solubility is given as 1.1*10^(-7) M.

Molarity (M) = moles of solute / volume of solution (in liters)

1.1*10^(-7) M = x moles / 1 L (since the volume is 1 liter)

Rearranging the equation, we get:
x moles = 1.1*10^(-7) M * 1 L

Step 2: Calculate the molar mass of calcium phosphate. The molar mass of calcium (Ca) is 40.08 g/mol, and the molar mass of phosphate (PO4) is 94.97 g/mol.

Molar mass of calcium phosphate (Ca3(PO4)2) = 3 * molar mass of calcium + 2 * molar mass of phosphate
= 3 * 40.08 g/mol + 2 * 94.97 g/mol

Step 3: Calculate the solubility in grams per liter.

Solubility (in g/L) = (number of moles) * (molar mass of calcium phosphate)

Plug in the values obtained in step 1 and step 2 into the formula and calculate.

For the second question about the unknown liquid, you are correct that you need to find the number of moles using the given information and then calculate the molar mass.

To find the number of moles, you can use the freezing point depression equation: ΔT = Kf * m * i, where ΔT is the change in freezing point, Kf is the freezing point depression constant, m is the molality of the solution, and i is the van 't Hoff factor (the number of particles produced by one formula unit of the solute when it dissociates in the solvent).

In this case, tert-butyl alcohol is the solvent, and the unknown liquid is the solute. The change in freezing point (ΔT) is given as 10.2 °C, the freezing point depression constant (Kf) is given as 9.10 °C/m, and the molality (m) can be calculated using the formula: m = moles of solute / mass of solvent (in kg).

The molality (m) can be calculated as follows:

m = (moles of solute) / (mass of solvent in kg)
= (0.807 g) / (11.6 g) * (1 kg / 1000 g)
= (0.807 / 11.6) * (1/1000) kg

Once you have calculated the molality, you can rearrange the freezing point depression equation to solve for the moles of the unknown liquid.

Finally, you can calculate the molar mass of the unknown liquid using the formula molar mass = mass in grams / number of moles.