one more chemistry problem :{?

[miss KT]

ok i know you all probably HATE me by now but i'm worried about on last problem

What is the Kb of a solvent if 6.23 g of Li3PO4 are dissolved in 100.0 g of the solvent. The boiling point of the solution was 72.3°C and the boiling point of the pure solvent is 70.8°C.

my answer -

dT= i * Kb * m
i= 3Li+ + PO4- = 4
dT= 72.3 C - 70.8 C = 1.5 C
m= mol/kg
(6.23 mol Li3PO4/1) * (1 mol Li3PO4/ 115.8 g LI3PO4)= 0.0537 mol Li3PO4
(0.0538 mol Li3PO4/0.100 kg)= 0.538m
Kb= dT/ i*m
Kb= 1.5 C/ 4 * 0.538
Kb= 0.69 C/m


again..am i right??

[the masked zorro]

u so fine u blow my mind
haha idk sorry

[Arkadev G]

:-{}{}}{}{}???????????????????. No

[Arkadev G]

:-{}{}}{}{}???????????????????. No