Question

Freezing point of a Solution of liqud unknown Freezing point depression: Trial #1. 0.0℃ (-3.4°C)= 3.4℃ Molality ot untnown slution, Mu Mass of Solvent (water) $4.3og-10, lag = 74.539 Moles of Solute = molality x masso solvent 0.1353 mol malar mass of unknown: tla0 x 1000 g 1000910 1Z9135. I k 135.59 132m

Molar mass determination by depression of freezing point lab I'm stuck on calculating the moles of solute.. How do I calculate it? Also can u please check if I've done everything else correctly.. The data I collected: Measured freezing point of pure water: 0.0 degrees Celsius Actual mass of solute used: 10.12g Freezing point of solution (observed): -3.4 Celsius Mass of solution: 84.7g

Answer 1

We know that ΔT _f = K_f x m
Where

ΔT _f = depression in freezing point

        = freezing point of pure solvent – freezing point of solution

        = 0.0 -(-3.4) oC

        = 3.4 ^oC

K _f = depression in freezing constant of water = 1.86 °C/m

m = molality of the solution

    = number of moles of solute / / weight of the solvent in Kg

    = ( mass / Molar mass ) / weight of the solvent in Kg

    = (10.12 g / M ) / (84.7g/1000(g/kg))

    = 119.55/M

Plug the values we get 3.4 = 1.86x(119.5/M)

                                   M = 65.5 g/mol

Therefore the molar mass of solute is 65.5 g/mol

So number of moles of solute, n = mass/molar mass

                                              = 10.12 g / 65.5(g/mol)

                                              = 0.155 mol