A student prepares a solution by dissolving 209 g of para...

A student prepares a solution by dissolving 2.09 g of para-dichlorobenzene (C6H4Cl2, = 147.0 g/mol) in 50.0 mL of cyclohexane. What is the freezing point (in °C) of the prepared solution? Pure cyclohexane has a freezing point of 6.47 °C, a freezing point depression constant of 20.0 °C/m, and a density of 1.25 g/mL. Freezing point = ______ °C

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00:01 So to solve for this, we are going to use the equation for the freezing point depression.

00:09 So we have freezing point depression is equivalent to a bantap factor of the solute multiplied by the molar freezing point depression constant, f, multiplied by molality.

00:26 So remember that malality is, this is moolute over.

00:30 We must have the solvent in kilograms.

00:34 So this is kilogram solvent.

00:36 So our unknown here is the freezing point, rather, of the solution.

00:45 So take note that the freezing point depression, tf, is equivalent to the freezing point of the pure solvent, minus the freezing point of the solution.

01:01 Okay, so this is our unknown, the freezing point of the solution.

01:03 So solving for the freezing point depression, delta tf is equivalent to.

01:09 The band -f factor for paradictor of benzene is just one.

01:16 Try to year one.

01:17 And then kf is 20 degrees celsius per molar, okay, with moles per kilogramm.

01:28 And then for molality, we have most absolute, so we have the mass, which is 2 .09.

01:34 Grams.

01:35 How do we convert this to moles? we're going to see the molar mass of our salute and that's equivalent to 147 .0 grams for every mole of the salute.

01:48 Divided by the mass of the substance kilogram solvent we have with tml, right? that's volume...

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