CHM 104 (General Chemist n Exam 1 on Chapters 11 and 13: Spring 2017, Essex coun
ID: 480379 • Letter: C
Question
CHM 104 (General Chemist n Exam 1 on Chapters 11 and 13: Spring 2017, Essex county lege. E. Aouad 16. Aliquid has an enthalpy of vaporization of30.8klfmol. At 269 K ithas a pressure of normal boiling point of this liquid? (R- 8.3 l4 JNK mol) a. 287 K b. 314 K. c. 269 K d. 253 K. e. 235 K vapor 107 mmHg. What is the 17. ich of the following statements isare CORRECT? solubility is as the concentration of solute in equilibrium with undissolved solute in a saturated solution. 2. If two liquids mix to an appreciable extent to form a solution, they are miscible. If two mix completely in any proportion to form a solution, the resulting solution is supersaturated. a. I only b. 2 only c. 3 only d. 1 and 2 e. 2 and 3 18. What mass of Namsos must be dissolved in 100.0 grams of water to lower the freezing point by 2.50°C? The freezing point depression constant, Kip, ofwater is -1.86 °Cm. Assume the van't Hoff factor for Na2so4 is2.ss. a. 3.77g b. 6.36 g c. 6.70 g e. 19.1 gExplanation / Answer
16) Ans= b - 314 K
Boiling point : The temperature at which vapor pressure of liquid is equal to atmospheric pressure.
Clasius-Claperyon equation is
In(P2/P1) = [Hvap/R] [(1/T1) - (1/T2)] -----Eq(1)
where R = universal gas constant = 8.314 J/K/mol
T1 = 269 K
T2 = ?
P1 = 107 mmHg
P2 = atmospheric pressure = 1 atm = 760 mmHg
Hvap = 30.8 kJ/mol = 30800 J/mol
Substitute all the values in eq (1) i.e.
In(P2/P1) = [Hvap/R] [(1/T1) - (1/T2)]
In (760 mmHg/107 mmHg) = [30800 J/mol /8.314 J/K/mol] [(1/269) - (1/T2)]
On simplification,
T2 = 314 K
Therefore,
Normal boiling point of the liquid = 314 K