(1a) The boiling point of chloroform CHCl3 is 61.70°C at 1 atmosphere. A nonvola
ID: 487206 • Letter: #
Question
(1a) The boiling point of chloroform CHCl3 is 61.70°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in chloroform is cholesterol. If 13.94 grams of cholesterol, C27H46O (386.6 g/mol), are dissolved in 214.5 grams of chloroform.
The molality of the solution is = m.
The boiling point of the solution is = °C.
(b) The freezing point of water H2O is 0.00°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is sucrose . If 12.97 grams of sucrose, C12H22O11 (342.3 g/mol), are dissolved in 156.7 grams of water ...
The molality of the solution is = m.
The freezing point of the solution is = °C.
(2a) The boiling point of water, H2O, is 100.000 °C at 1 atmosphere. Kb(water) = 0.512 °C/m In a laboratory experiment, students synthesized a new compound and found that when 10.69 grams of the compound were dissolved in 281.3 grams of water, the solution began to boil at 100.057 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?
= g/mol
(b) The freezing point of water, H2O, is 0.000 °C at 1 atmosphere. Kf(water) = 1.86 °C/m In a laboratory experiment, students synthesized a new compound and found that when 11.22 grams of the compound were dissolved in 200.4 grams of water, the solution began to freeze at -0.578 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound?
= g/mol
(c) The boiling point of chloroform, CHCl3, is 61.700 °C at 1 atmosphere. Kb(chloroform) = 3.67 °C/m In a laboratory experiment, students synthesized a new compound and found that when 12.26 grams of the compound were dissolved in 290.9 grams of chloroform, the solution began to boil at 62.100 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?
= g/mol
Explanation / Answer
Q1.
The boiling point of chloroform CHCl3 is 61.70°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in chloroform is cholesterol. If 13.94 grams of cholesterol, C27H46O (386.6 g/mol), are dissolved in 214.5 grams of chloroform.
a)
The molality of the solution is = m
m = mol of solute / kg solvnet = (13.94 /386.6) / (214.5 /1000) = 0.1681
b)
The boiling point of the solution is = °C. --> apply colligative properties
dTb = Kb*m = 0.1681*3.88 = 0.652228
Tb normal = 61.2
Tb mix = 61.2+ 0.652228 = 61.852 °C
Q2.
The freezing point of water H2O is 0.00°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is sucrose . If 12.97 grams of sucrose, C12H22O11 (342.3 g/mol), are dissolved in 156.7 grams of water ...
a)
The molality of the solution is = m.
molality = mol of solute / kg solvent = (mass/MW) / (kg solvent) = (12.97 /342.3 ) /(156.7 /1000) = 0.24180
The freezing point of the solution is = °C.
Tf water= 0°C
dTf mix = -Kf*m* = -1.86*0.24180 = -0.449748°C
Tf mix = 0-0.449748 = -0.449748°C
Please follow Chegg's Guideline for multiple questions. We are only allowd to answer to one single questions per set of Q&A. Please post all other questions in another set of Q&A.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.