Use the graph of vapor pressure to determine the normal boiling point of CHCI_3.

Question

Use the graph of vapor pressure to determine the normal boiling point of CHCI_3. A) 19 degree C B) 52 degree C C) 60 degree C D) 64 degree C E) 70 degree C The vapor pressure of a liquid in a closed container depends upon A) he amount of liquid. B) the surface area of the liquid. C) the volume of the container. D) the temperature. E) None of the above. The molar enthalpy of vaporization of hexane (C_6H_14) is 28.9 k/mol, and its normal boiling point is 68.73 degree C. What is the vapor pressure of hexane at 25 degree C? A) 171 torr B) 4.44 torr C) 117 torr D) 3370 torr E) 759 torr The normal boiling point of methanol (CH_3OH) is 64.6 degree C. Given that the vapor pressure of methanol is 75.0 torr at 15.2 degree C, calculate the molar enthalpy of vaporization of methanol. A) 0.383 kJ/mol B) 3.00 kJ/mol C) 38.0 kJ/mol D) 27.5 kJ/mol E) 74.7 kJ/mol Of the pair of compounds given, which would have the stronger intermolecular forces of attraction? CH_4 or CH_3OH Given that the heat of vaporization of diethyl ether is 0 kJ/mol and the vapor pressure of diethyl ether is 440 torr at 20. degree C, calculate the normal boiling point of diethyl ether.

Explanation / Answer

15. Boiling point is the temperature at which vapour pressure is equals to atmospheric pressure. the atmospheric pressure is 760 mm of Hg and the corresponding temperature is 64 oC. hence, the answer is D) 64 oC.

16. The vapour pressure of a liquid in a closed container depends upon: B) the surface area of the liquid. Because in a closed container there is an equlibrium exist between the liquid phase and vapour phase, more the amount if vapour formed more is the vapour pressure. It depends upon the surface area of the liquid. more the surface area more the amount of vapour formed.

Hence the answer is: B) the surface area of the liquid.

17. The Clausius-Clapeyron Equation:

ln (P1 / P2) = - (H / R) (1/T1 - 1/T2)

Given data:

P1 = ? = x, say, T1 = 25.00 °C = 298.15 K, P2 = 760.0 mmHg and T2 = 68.73 °C = 341.88 K

Putting the values in equation :

ln (x / 760) = - (28900 / 8.31447) (1/298.15 minus 1/341.88)

ln (x / 760) = - 1.4912

x /760 = 0.2251

x = 171 torr

Answer: A) 171 torr

18.

The Clausius-Clapeyron Equation:

ln (P1 / P2) = - (H / R) (1/T1 - 1/T2)

Given data:

P1 = 760 torr, T1 = 64.6 °C = 337.75 K, P2 = 75 torr and T2 = 15.2 °C = 288.35 K

Putting the values in equation : ln (760 / 75) = - (H / 8.314) (1/337.75 - 1/288.35)

or, 2.32 =  - H (-0.000061)

or, H = 2.32/0.000061 = 38,026.55 J/mol = 38 kJ/mol

Answer: C) 38.0 kJ/mol

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