A solution containing 1.58 g of unknown molecular compound dissolved in 15.0 g o

Question

A solution containing 1.58 g of unknown molecular compound dissolved in 15.0 g of water freezes at -1.25 oC. KF for water = 1.86 oC/m. a) What is the solute and solvent in the solution? b) What is the freezing point depression (Tf)? c) What is the molality (m) of solution in mol/kg? (Tf = i x Kf x m) Show your calculations. d) How many moles of unknown substance is used in the solution? Show your calculations. e) What is the experimental molecular weight (or mass) of the unknown substance in g/mol? Show your calculations.

Explanation / Answer

a)

solute = least amount, in this case, the 1.58 of unkown

solvent = the most amount, this is water, typically

b)

find dTf -->

Tf normal = 0°C

dTf = -1.25°C

c)

molality from:

dTf = -m*Kf

m = dTf/Kf = 1.25/1.86 = 0.6720

d)

moles of subtance

molality = mol solute / kg solvent

mol solute = molality * kg = 0.6720 * 15*10^-3 = 0.01008

e)

MW = mass/mol = 1.58/0.01008

MW = 156.74 g/mol

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.