Molar mass determination by depression of freezing point lab I\'m stuck on calcu

Question

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

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

Explanation / Answer

We know that T f = Kf x m
Where

T 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

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