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The molar volume of liquid benzene (density = 0.877 g mL-1) increases by a factor of 2750 as it vaporizes at 20°C and that of liquid toluene (density = 0.867 g mL-1) increases by a factor of 7720 at 20°C. A solution of benzene and toluene at 20°C has a vapour pressure of 46.0 Torr. Find the mole fraction of benzene in the vapour above the solution.

Shane Macguire , 10 Years ago
Grade upto college level
anser 2 Answers
Deepak Patra

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
(i)Volume = No. of moles * molar mass/density
(ii) PV = nRT or P = nRT/V
Volume of 1 mole of liq. Benzene = 78/0.877
Volume of 1 mole of toluene. 92/0.867
In vapour phase,
At 20oC, for 1 mole of benzene,
Volume = 1 **78 *2750/0.877 = 244583.80 mL
= 244.58 L
Similarly for 1 mole of toluene,
Volume = 1 *92/0.867 * 7720
= 819192.61 mL = 819.199 L
As we know that, PV = nrt
For benzene, PoB = nRT/V = 1 *0.0821 *293/244.58 atm
= 0.098 atm
For toluene, PoT = nRT/V = 1 *0.0825 *293/819.19 atm = 0.029 atm
P = PoB.XB + PoT.XT
∵XB + XT = 1
∴ XT= 1 - XB
P = PoB.XB + PoT (1 – XB)
Total vapour pressure = 46 torr = 46/760 = 0.060 atm
Thus, 0.060 = 0.098 XB + 0.029 (1 - XB)
⇒ 0.060 = 0.098 XB + 0.029 – 0.029 XB
⇒ 0.031 = 0.069 XB
∴ XB = 0.031/0.069 = 0.45 (in liquid phase)
XB + XT = 1
XT = 1 – 0.45 = 0.55 (in liquid phase)
Also, P’B = PoB.XB = PX’B
So, 0.098 * 0.045 = 0.060 * X’B
X’B = 0.098 * 0.45/0.060 = 0.735 (in gas phase)

Thanks
Deepak patra
askIITians Faculty

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the answer to your question
 
(i)Volume = No. of moles * molar mass/density
(ii) PV = nRT or P = nRT/V
Volume of 1 mole of liq. Benzene = 78/0.877
Volume of 1 mole of toluene. 92/0.867
In vapour phase,
At 20oC, for 1 mole of benzene,
Volume = 1 **78 *2750/0.877 = 244583.80 mL
= 244.58 L
Similarly for 1 mole of toluene,
Volume = 1 *92/0.867 * 7720
= 819192.61 mL = 819.199 L
As we know that, PV = nrt
For benzene, PoB = nRT/V = 1 *0.0821 *293/244.58 atm
= 0.098 atm
For toluene, PoT = nRT/V = 1 *0.0825 *293/819.19 atm = 0.029 atm
P = PoB.XB + PoT.XT
∵XB + XT = 1
∴ XT= 1 - XB
P = PoB.XB + PoT (1 – XB)
Total vapour pressure = 46 torr = 46/760 = 0.060 atm
Thus, 0.060 = 0.098 XB + 0.029 (1 - XB)
⇒ 0.060 = 0.098 XB + 0.029 – 0.029 XB
⇒ 0.031 = 0.069 XB
∴ XB = 0.031/0.069 = 0.45 (in liquid phase)
XB + XT = 1
XT = 1 – 0.45 = 0.55 (in liquid phase)
Also, P’B = PoB.XB = PX’B
So, 0.098 * 0.045 = 0.060 * X’B
X’B = 0.098 * 0.45/0.060 = 0.735 (in gas phase)
 
Thanks and regards,
Kushagra

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