a sample of 15.00 g of an unkown metal chloride, MCl2 is dissolved in 75.0 mL water. the boiling point of the resulting solution is measured at 102.37 degrees C. what is the identity of the metal M.

Answer 1

I assume MCl2 is an ionic chloride so i, the van't Hoff factor, is 3.
delta T = i*Kb*m
You know delta T, Kb, and i (= 3). Solve for molality, m.
Then m = mols/kg solvent.
You know kg solvent and m, solve for mols.
Then mols = grams/molar mass
You know mols and grams, solve for molar mass.

Finally, knowing molar mass MCl2 and subtrating 2*Cl, you get atomic mass M. Identify that form the periodic table.

Answer 2

what's the T Kb what would the molality be plugged in for the equation

Answer 3

delta T = 102.37-100 = ?
i = 3
Kb = 0.512
m = ?

Answer 4

m= 75 mL of water 0.001kg water= o.o75 of warer.
2.37 c = 3*0.512 c/m*m
m=1.54297

to get moles
1.59297*0.075/1=0.119473 mol

molar mass
15/ 0.119473= 125.551

125.551g -2* 35.45= 54.6514 making it manganese chloride would that be correct ?

Answer 5

Close but I think you have a math error.

Here is your work.
m= 75 mL of water 0.001kg water= o.o75 of warer.
2.37 c = 3*0.512 c/m*m
m=1.54297
You are ok to here

to get moles
1.59297*0.075/1=0.119473 mol
I think you mistyped (transferred from the step above incorrectly) which makes the mols slightly out of place. That makes the final answer not quite right. The procedure looks ok though; therefore, I think you will be ok if you go back through and correct thos steps. I think the final answer comes out closer to Ni than Mn but check that out.

molar mass
15/ 0.119473= 125.551

125.551g -2* 35.45= 54.6514 making it manganese chloride would that be correct ?