What was the initial boiling point if 44.20 g NaCl was added to 35.00 kg of...
- What was the initial boiling point if 44.20 g NaCl was added to
35.00 kg of an unknown solvent, making the final boiling point of
the solution 130.00℃? (use K = 2.00 ℃ᆞKg/mol)
- 124.95℃
- 135.05 ℃
- 46.80℃
- 129.5℃
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What was the initial boiling point if 44.20 g NaCl was added to
35.00 kg of...
21. 13.5 g of NaCl is dissolved in 100.0 g of H20. The boiling point of...
21. 13.5 g of NaCl is dissolved in 100.0 g of H20. The boiling point of the water after the salt is added is The boiling point elevation constant for water: Ko = 0.512 °C/m. A) 102.4°C B) 101.2°C C) 104.5°C D) 108.4°C 22. A process has an equilibrium constant K of 2.5 x 10 at 300 K. What is AGⓇ at that temperature? A) 19.5 kJ/mol B) 25 kJ/mol C) -19.5 kJ/mol D) 75 kJ/mol 23. In heterogeneous catalysis,...
21. 13.5 g of NaCl is dissolved in 100.0 g of H20. The boiling point of...
21. 13.5 g of NaCl is dissolved in 100.0 g of H20. The boiling point of the water after the salt is added is The boiling point elevation constant for water: Kp = 0.512 °C/m. A) 102.4° C B) 101.2°C C) 104.5°C D) 108.4°C 22. A process has an equilibrium constant K of 2.5 x 10' at 300 K. What is AG° at that temperature? A) 19.5 kJ/mol B) 25 kJ/mol C) -19.5 kJ/mol D) 75 kJ/mol 23. In heterogeneous...
A Review Constants Periodic Table The changes in boiling point (AT) or freezing point (AT) in...
A Review Constants Periodic Table The changes in boiling point (AT) or freezing point (AT) in degrees Celsius from a pure solvent can be determined from the equations given here, respectively: AT) = m x K = moles of solute XK K. kilograms of solvent Since pure water boils at 100.00 °C, and since the addition of solute increases boiling point, the boiling point of an aqueous solution, Th, will be T - (100.00+AT) 'C Since pure water freezes at...
a) If 1.95 mol of calcium chloride is added to 2.30 kg of water then what...
a) If 1.95 mol of calcium chloride is added to 2.30 kg of water then what is the freezing point lowering and the freezing point of the solution (Kf = 1.86 ºC/m)? b) If 0.500 mol of NaCl is added to 500 mL of water then what is the boiling point elevation and the boiling point of the solution (Kb = 0.52 ºC/m)?
Which solution has the higher boiling point: 24.0 g C2H6O2 in 0.500 kg of H2O or...
Which solution has the higher boiling point: 24.0 g C2H6O2 in 0.500 kg of H2O or 20.0 g NaCl in 0.500 kg of H2O? A. C2H6O2 has the higher boiling point. B. NaCl has the higher boiling point.
Estimate the boiling point elevation of 0.39 m NaCl(aq), assuming it undergoes complete dissociation. The boiling...
Estimate the boiling point elevation of 0.39 m NaCl(aq), assuming it undergoes complete dissociation. The boiling point constant of water is 0.51 K · kg/mol. Answer in units of ◦C.
If 1 kg of solvent (water) and 10 g of a solute (unknown) were mixed and the boiling point determined to be 101.8 degree...
If 1 kg of solvent (water) and 10 g of a solute (unknown) were mixed and the boiling point determined to be 101.8 degrees Celsius, calculate the molecular weight of the solute. Assume the water boiled at 100 degrees Celsius before solute was added.
Review Constants Periodic Table The changes in boiling point (AT) or freezing point (AT) in degrees...
Review Constants Periodic Table The changes in boiling point (AT) or freezing point (AT) in degrees Celsius from a pure solvent can be determined from the equations given here, respectively: Value Units moles of solute AT = mx Kb = 7 Submit kilograms of solvent XRb moles of solutex Kf Part B AT: = mx Kf = kilograms of solvent where m is the molality of the solution, and K and K the boiling-point-elevation and freezing-point-depression constants for the solvent,...
What is the boiling point of a solution produced by adding 650 g of cane sugar...
What is the boiling point of a solution produced by adding 650 g of cane sugar (molar mass 342.3 g/mol) to 1.7 kg of water? For each mole of nonvolatile solute, the boiling point of 1 kg of water is raised 0.51 ∘C.
5. When 4.50 g of a substance are dissolved in 125 g of CCl, (M-153.8 g/mol),...
5. When 4.50 g of a substance are dissolved in 125 g of CCl, (M-153.8 g/mol), the freezing point of the solution is depressed by 3.9 K. KCCl) 30 and K,(CC4) 4.95 K kg mol 0.64 K] a) What is the boiling point elevation of the solution? b) By what fraction is the vapor pressure of the solvent reduced? 1
21. 13.5 g of NaCl is dissolved in 100.0 g of H20. The boiling point of the water after the salt is added is The boiling point elevation constant for water: Ko = 0.512 °C/m. A) 102.4°C B) 101.2°C C) 104.5°C D) 108.4°C 22. A process has an equilibrium constant K of 2.5 x 10 at 300 K. What is AGⓇ at that temperature? A) 19.5 kJ/mol B) 25 kJ/mol C) -19.5 kJ/mol D) 75 kJ/mol 23. In heterogeneous catalysis,...
21. 13.5 g of NaCl is dissolved in 100.0 g of H20. The boiling point of the water after the salt is added is The boiling point elevation constant for water: Kp = 0.512 °C/m. A) 102.4° C B) 101.2°C C) 104.5°C D) 108.4°C 22. A process has an equilibrium constant K of 2.5 x 10' at 300 K. What is AG° at that temperature? A) 19.5 kJ/mol B) 25 kJ/mol C) -19.5 kJ/mol D) 75 kJ/mol 23. In heterogeneous...
A Review Constants Periodic Table The changes in boiling point (AT) or freezing point (AT) in degrees Celsius from a pure solvent can be determined from the equations given here, respectively: AT) = m x K = moles of solute XK K. kilograms of solvent Since pure water boils at 100.00 °C, and since the addition of solute increases boiling point, the boiling point of an aqueous solution, Th, will be T - (100.00+AT) 'C Since pure water freezes at...
Review Constants Periodic Table The changes in boiling point (AT) or freezing point (AT) in degrees Celsius from a pure solvent can be determined from the equations given here, respectively: Value Units moles of solute AT = mx Kb = 7 Submit kilograms of solvent XRb moles of solutex Kf Part B AT: = mx Kf = kilograms of solvent where m is the molality of the solution, and K and K the boiling-point-elevation and freezing-point-depression constants for the solvent,...
5. When 4.50 g of a substance are dissolved in 125 g of CCl, (M-153.8 g/mol), the freezing point of the solution is depressed by 3.9 K. KCCl) 30 and K,(CC4) 4.95 K kg mol 0.64 K] a) What is the boiling point elevation of the solution? b) By what fraction is the vapor pressure of the solvent reduced? 1
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