owLv2 | Online teaching and ; eAssignment |a search 110% Review Topics Reference

Question

owLv2 | Online teaching and ; eAssignment |a search 110% Review Topics References Use the References to access important values if needed for this question. The freezing point of benzene. CoHs, is 5.500 °C at 1 atmosphere. K(benzene) 5.12 °Cm In a laboratory experiment, students synthesized a new compound and found that when 10.69 grams of the compound were dissolved in 212.5 grams of benzene, the solution began to freeze at 4.834 °C The compound was also found to be nonvolatile and a non- electrolyte What is the molecular weight they determined for this compound g mol Hiintire 11:32 PM

Explanation / Answer

Ans. Step 1: Let the molar mass of unknown = X g/mol

Now,

            Moles of compound = Mass / MW = 10.69 g / (X g/ mol) = (10.69 / X) mol

Molality of solution = Moles of Solute / Mass of solvent in kg

= (10.69 / X) mol / 0.2125 kg = (50.306 / X) m

# Depression in freezing point, dTf =

Freezing point of pure solvent – Freezing point of solution

= 5.5000C – 4.8340C = 0.6660C

# Step 2: Depression in freezing point of the solution is given by-

                        dTf = i Kf m             - equation 1

            where, i = Van’t Hoff factor. [ i = 1, given]

                        Kf = molal freezing point depression constant (of water) = 5.120C / m

                        m = molality of the solution

                        dTf = Freezing point of pure solvent – Freezing point of solution

Putting the values in equation 1-

            0.6660C = 1 x (5.12 0C m-1) x (50.306 / X) m

            Or, 0.6660C = 257.566720C / X

            Or, X = 257.566720C / 0.6660C

            Hence, X = 392.507

Therefore, molar mass of sugar = X g/ mol = 392.507 g/ mol

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