Determine the freezing point of 0.368 kg of h2o with 11.85 g of c2h5oh where the kf is 1.86 c/m.

We will use the formula for freezing point depression :

but first, we need to get the molality m of the solution:

- molality m = moles of C2H5OH / mass of water Kg

 when moles of C2H5OH = mass of C2H5OH/ molar mass of C2H5OH

                                            = 11.85 g / 46 g/mol
                                 
                                           = 0.258 moles

and when we have the mass of water Kg = 0.368 Kg

so, by substitution on the molality formula:

∴ molality m = 0.258 moles / 0.368 Kg

                     = 0.7 mol/Kg

and when C2H5OH is a weak acid so, there is no dissociation ∴ i = 1

and when Kf is given = 1.86 C/m

so by substitution on ΔTf formula:

when ΔTf = i Kf m

∴ ΔTf = 1 * 1.86C/m * 0.7mol/Kg

          = 1.302 °C

Victor's sense of being challenged to understand the structure of the body.

Explanation:

Just took it.