1. Carbon tetrachloride has a vapor pressure of 213 torr at 40.°C and 915 torr a

Question

1. Carbon tetrachloride has a vapor pressure of 213 torr at 40.°C and 915 torr at 83°C. What is the normal boiling point of CCl4? Boiling point = ___ C 2. Diethyl ether was one of the first chemicals used as an anesthetic. At 34.6°C, diethyl ether has a vapor pressure of 760. torr, and at 18.1°C, it has a vapor pressure of 407 torr. What is the H of vaporization for diethyl ether? H of vaporization = ___ kJ/mol 3.An aqueous antifreeze solution is 23.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 . Calculate the molality, molarity, and mole fraction of the ethylene glycol Molality = ___mol/kg Molarity = ___ mol/L Mole fraction =

Explanation / Answer

1. Carbon tetrachloride has a vapor pressure of 213 torr at 40.°C and 915 torr at 83°C. What is the normal boiling point of CCl4? Boiling point = ___ C

Solution :- Lets first calculate the heat of vaporization of the CCL4 using the Clausius Clayperon equation.

P1 = 213 torr , T1 = 40+273 = 313 K

P2 = 915torr , T2 = 83+273 = 356 K

Ln[P2/P1] = delta H vap / R [(1/T1)-(1/T2)]

Ln[915/213]= delta H vap / 8.314 J per mol K [(1/313)-(1/356)]

1.4576 = [delta H vap / 8.314 J per mol K] [0.000386]

1.4576*8.314 J per mol K / 0.000386 = delta H vap

31395 J per mol K= delta H vap

Now using the delta H vap we can calculate the normal boiling point

At the normal boiling point the vapor pressure is 760 torr

So lets assume T1 = 313 K and P1 = 213 torr

T2 = ? , P2 = 760 torr

Ln[P2/P1] = delta H vap / R [(1/T1)-(1/T2)]

Ln[760/213]= 31395 J per mol / 8.314 J per mol K [(1/313)-(1/T2)]

1.272 = 3776 [0.003195 – (1/T2)]

1.272/3776 = 0.003195 – (1/T2)

0.0003369 = 0.003195 – (1/T2)

0.0003369-0.003195 = -1/T2

-0.002858 = -1/T2

T2= -1/-0.002858

T2 = 350 K

350 K – 273 = 77 C

So the normal boiling point is 77 C for

2. Diethyl ether was one of the first chemicals used as an anesthetic. At 34.6°C, diethyl ether has a vapor pressure of 760. torr, and at 18.1°C, it has a vapor pressure of 407 torr. What is the H of vaporization for diethyl ether? H of vaporization = ___ kJ/mol

Solution :-

P1 = 760 torr , T1 = 34.6 C + 273 = 307.6 K

P2 = 407 torr , T2 = 18.1 C +273 = 291.1 K

Delta H vap = ?

Ln[P2/P1] = delta H vap / R [(1/T1)-(1/T2)]

Ln[407/760]= delta H vap / 8.314 J per mol K [(1/307.6)-(1/291.1)]

-0.6245 = [delta H vap / 8.314 J per mol K] [-0.00018]

-0.6245 * 8.314 J per mol K / -0.00018 K = Delta H vap

28845 J per mol = delta H vap

28845 J per mol * 1 kJ / 1000 J = 28.84 kJ/mol

So the enthalpy of vaporization is 28.84 kJ/mol

3.An aqueous antifreeze solution is 23.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 . Calculate the molality, molarity, and mole fraction of the ethylene glycol Molality = ___mol/kg Molarity = ___ mol/L Mole fraction =

Solution :-

Lets assume we have 1 liter solution

Mass of solution = volume * density

Mass of solution = 1000 ml * 1.05 g per ml = 1050 g

Lets find the mass of solute

Mass of solute = 1050 g * 23.0 % / 100 % = 241.5 g

Mass of solvent = mass of solution - mass of solute

                           = 1050 g – 241.5 g

                           = 808.5 g

Now lets calculate the moles of solute and solvent

Moles = mass / moalr mass

Moles of ethylene glycol = 241.5 g / 62.07 g per mol = 3.891 mol

Moles of water = 808.5 g / 18.015 g per mol = 44.88 mol

Molality = moles / volume in liter

                = 3.891 mol / 1 L

                = 3.891 M

Molality = moles / kg solvent

                = 3.891 mol / 0.8085 kg

                = 4.813 m

Mole fraction = moles of solute / total moles

                       = 3.891 mol / (3.891 mol + 44.88 mol )

                       = 0.0798

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