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The molar volume of liquid benzene increases by a factor of 2750 as it vaporizes at 20 C. At 27 C when a non volatile solute is dissolved in 54.6 cm3 of benzene vapour pressure of this solution is found to be 98.88mmHg. Calculate the freezing point of the solution Given Enthalpy of vapourisation of benzene = 394.57J/g Enthalpy of fusion of benzene(l) = 10.06KJ/mol Molal depression constant for benzene= 5.0 K Kg/ Mol

The molar volume of liquid benzene increases by a factor of 2750 as it vaporizes at 20 C. At 27 C when a non volatile solute is dissolved in 54.6 cm3 of benzene vapour pressure of this solution is found to be 98.88mmHg. Calculate the freezing point of the solution
 
Given Enthalpy of vapourisation of benzene = 394.57J/g
Enthalpy of fusion of benzene(l) = 10.06KJ/mol
Molal depression constant for benzene= 5.0 K Kg/ Mol

Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
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)

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