Help me solve these part please, and explain for Irial 2 ol test tube Mass of cy

Question

Help me solve these part please, and explain for Irial 2 ol test tube Mass of cyclohexane Freezing point of pure cyclohexane Mass of naphthalene and paper Mass of paper Total mass of naphthalene in solution 19 315 Freezing point of solution Freezing-point depression, AT Molality of solution, m Molal freezing-point constant Average k B. Molecular Weight of an Unknown Solute /12 Trial 2 Trial 1 10 7.83) Mass of test tube and cyclohexane Mass of test tube Mass of solvent Mass of sample and paper Mass of paper Total mass of solid sample in solution Freezing point of solvent (from part A) Freezing point of solution Freezing-point depression, ar, Apparent molality of solution Grams of solute per kg of solvent Molecular weight of solute (grams per mole) Average molecular weight of solute 92 611) O , 3 3039- 129

Explanation / Answer

Ans. #Part B. Trial 1:

Depression in freezing point, dTb =

Freezing point of pure solvent – Freezing point of solution

= 7.00C – 3.00C

= 4.00C

Freezing point depression, dTf is given by-

            dTf = i Kf m               - equation 1

            where, i = Van’t Hoff factor. [i = 1 for non-electrolyte solute].

                        Kf = molal freezing point depression constant of cyclohexane = 20.00C / m

                        m = molality of the solution

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

Since no information is provided for van’t Hoff factor, it’s assumed that the solute is non-electrolyte or i =1.

Let the molality of the resultant solution be m.

Putting the values in equation 1-

            4.00C = 1 x (20.00C / m) x m

            Or, m = 4.00C / (20.00C / m) = 0.20 m

Therefore, calculated molality of resultant solution = 0.20 m

# Determination of molar mass of solute:

Mass of solvent (cyclohexane) = 10.141 g = 0.010141 kg

# Let the molar mass of solute be X g/ mol

Now, moles of solute = Mass of solute / Molar mass

                                    = 3.03 g / (X g/mol)

                                    = (3.03 / X) mol

Now,

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

                                                = (3.03 / X) mol / 0.010141 kg

                                                = (298.787 / X) m

# From first part above, we have calculated molality of solution, m = 0.20 m

Now, both the terms of molality of the solution must be equal as-

            0.20 m = (298.787 / X) m

            Or, X = 298.787 / 0.20 = 1493.935

Therefore, molar mass of the solute = X g/mol = 1493.935 g /mol

#Part B. Trial 2:

Since dTf, mass of solute and mass of solvent remains the same as trial 1, the calculated molar mass of solute remains to be as in 1493.935 g /mol trial 1.

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