What is the freezing point of automobile radiator fluid prepared by mixing equal

Question

What is the freezing point of automobile radiator fluid prepared by mixing equal volumes of ethylene glycol and water at a temperature where the density of ethylene glycol is 1.114 g/mL an the density of water is 1 g/mL. When 20.0 grams of an unknown non electrolyte compound are dissolved in 500.0 grams of benzene, the freezing point of the resulting solution is 3.77 degreeC. The freezing point of pure benzene is 5.444 degreeC and the K, for benzene is 5.12 degreeC/m. What is the molar mass of the unknown compound? A 1.60-e sample of a mixture of naphthalene (C_10H_8) and anthracene (Cl_4H_10) is dissolved in 20.0 g benzene (C_6H_6). The freezing point of the solution is found to be 2.81 degreeC. What is the composition as mass percent of the sample mixture?

Explanation / Answer

Solution :-

Q8) Given data

Lets assume we have 100 ml water and 100 ml wthylene glycol

Then mass of water = 100 g and mass of ethylene glycol = 100 ml * 1.114 g per ml = 111.4 g

Lets calculate the moles of ethylene glycol

Moles = mass/ molar mass

            = 111.4 g / 62.07 g per mol

           = 1.795 mol

Now lets calculate the molality of the solution

Molality = moles / kg solvent

               =1.795 mol /0.100 kg

               = 17.95 m

Now lets calculate the change in freezing point

Delta Tf = Kf* m

Delta Tf = 1.86 C per m * 17.95 m

              = 33.4 C

Freezing point of solution = freezing point of solvent - delta Tf

                                               = 0 C – 33.4 C

                                               = -33.4 C

Q9) Lets first calculate the molality of the solution

Change in freezing point = 5.444 C – 3.77 C = 1.674 C

Delta Tf = Kf* m

Delta Tf / Kf = m

1.674 C / 5.12 C per m = m

0.327 m = m

Now lets calculate the moles of the solute

Moles = molality * kg solvent

          = 0.327 *0.500 kg

         = 0.1635 moles

Now lets calculate the molar mass of the solute

Molar mass = mass / moles

                     = 20 g / 0.1635 mol

                    = 122.3 g per mol

So the molar mass of solute = 122.3 g /mol

Q10) lets first calculate the molality of the solution

Change in freezing point = 5.455 C – 2.81 C = 2.645 C

Delta Tf = Kf * m

2.645 C = 5.065 c per m * m

2.645 C / 5.065 c per m = m

0.5222 m = m

Now lets calculate the total moles of the solute

Moles = moality * kg solvent

          = 0.5222 mol per kg * 0.020 kg

         = 0.010444 mol solute

Now lets assume moles of naphthalene = x

Then moles of anthracene = (1-x)

Total mass = 1.60 g

Now l;ets find the mass of the each solute

(Moles of naphthalene * molar mass )+(moles of anthracene* molar mass ) = 1.60 g

(x*128.17 g)+((0.010444-x)*178.23 g) = 1.60 g

Solving for x we get x= 0.0052

So moles of naphthalene = 0.0052 mol

So mass of napthelene = 0.0052 mol * 128.17 g per mol = 0.666 g

Mass of anthracene = 1.60 g – 0.666 g = 0.934 g

% of naphthalene = (0.666 g / 1.60 g)*100 % = 41.62 %

% of anthracene = 100 % - 41.62 % = 58.38 %

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.