1.) The common laboratory solvent diethyl ether (ether) is often used to purify
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Question
1.) The common laboratory solvent diethyl ether (ether) is often used to purify substances dissolved in it. The vapor pressure of diethyl ether, CH3CH2OCH2CH3, is 463.57 mm Hg at 25 °C. The molecular weight of diethyl ether, CH3CH2OCH2CH3, is 74.12 g/mol.
In a laboratory experiment, students synthesized a new compound and found that when 80.82 grams of the compound were dissolved in 262.6 grams of diethyl ether, the vapor pressure of the solution was 452.03 mm Hg. The compound was also found to be nonvolatile and a nonelectrolyte.
What is the molecular weight of this compound?
2.)A 1.60-g sample of a mixture of naphthalene (C10H8) and anthracene (C14H10) is dissolved in 20.0 g benzene (). The freezing point of the solution is 2.68°C. What is the composition as mass percent of the sample mixture? The freezing point of benzene is 5.51°C and is 5.12°C·kg/mol.
3)An antiseptic solution contains hydrogen peroxide, H2O2, in water. The solution is 0.567 m H2O2. What is the mole fraction of hydrogen peroxide?
Explanation / Answer
Vapor pressure of component (diethyl ether )in solution= its mole fraction* Pure component vapor pressure
452.03 =mole fraction* 463.57
Mole fraction of diethyl ether = 452.03/463.57 =0.975
Masses : Diethyl ether = 262.6 Unknown compoind = 80.82
Molecular weights : diethyl ether = 74. Molecular weight of unknown compound= M ( let us say)
Moles Diethyl ether= 262.6/74=3.55 Unknown compoind= 80.82/M
Mole fraction of diethyl ether = Moles of diethyl ether/ total moles= 3.55/(3.55+80.82/M) =0.975
3.55= 0.975*(3.55+80.82/M)
3.55*(1-0.975)= 0.975*80.82/M
0.08875= 0.975*80.82/M
M= 0.975*80.82/0.08875=887.88 g/mole
b) Depression in freezing point = -2.68=5.51=2.83 deg.c= kf*m
m= Molality , kf = Freezing point depression constant
2.83= m*5.12
m= 2.83/5.12 =0.53
molality= moles of solute kg of solvent
1 kg solvent contains 0.53 mole of solute ( in this case Napthalene and Anthracene)
20 gm =20/1000 kg of Benzene solvent contains 0.02*0.53 mole of solute= 0.0106
Total mass of mixture =1.6
let x = mole of Napthalene in the mixture and y= moles of Anthracene in the mixtute
hence x+y= 0.0106, x =0.0106-y (1)
mole = mass/ Molecular weight , Molecular weights : Napthalene : 128 Anthracene : 178
Mass of Napthalene = x*128, mass of anthracene= y*178
x*128+ y*178= 1.6
from !
(0.0106-y)*128+178y= 1.6, 1.36+y*(178-128)= 1.6
50y= 0.24, y=0.24/50=0.0048 x= 0.0106-0.0048=0.0058
moles of Napthalene= 0.0058 , mass of napthalene= 0.0058*128= 0.74gm and Anthracene= 1.6-0.74=0.86gm
Mass percentages : Napthalene : 100*0.74/1.6 =46.25% and Anthracene= 100-46.25=53.75%
c) the solution is 0.567 molall
that is 0.567 moles in 1 kg of solvent =1000 gms of solvent water
Moles of solvent water = 1000/18=55.56 moles
total moles= 0.567( H2O2)+ 55.56 (moles of water)=56.127
Mole fraction of H2O2= 0.567/56.127= 0.010
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