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D. Freezing Point Calculation Directions: Study the sample problem and do the task given. Task to do in Sample problem with solution Calculate the freezing point of a solution repared by dissolving 27,56 g of glucose (C6H1206) into 125 g of water. Ethylene glycol (C2H6O2) is a molecular compound that is used in many commercial anti-freezes. An aqueous solution of ethylene glycol is used in vehicle radiators to prevent the water in the radiator from freezing by lowering its freezing point. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500. g of water. Step 1: List the known quantities and plan the problem. Step 1: Kf (H2O) = -1.86 °C/m Known: mass of C2H6O2 = 400. g molar mass of C2H6O2 = 62.08 g/mol mass of H20 = 500. g = 0.500 kg Kf (H20) = -1.86 °C/m *Kf- is the constant freezing point coy nso ted Unknown: Tf of solution = ? °C *Tf- is the total freezing point This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression. Step 2: e er Step 2: Solve. 0 mal 400. g C2H6O2×1 mol C2H6O2÷62.08 g C2H6O2=6.44 mol C2H6O2 6.44 mol C2H602÷0.500 kg H2O=12.9 m C2H.O2 ATf = Kf xm = -1.86°C/m x 12.9 m = -24.0°C -24.0°C svio ents Tf %3D The normal freezing point of water is 0.0°C. Therefore, since the freezing point decreases by 24.0°C, the freezing point of the solution is -24.0°C. 2enE CHAVEE HER O 0f

D. Freezing Point Calculation Directions: Study the sample problem and do the task given. Task to do in Sample problem with solution Calculate the freezing point of a solution repared by dissolving 27,56 g of glucose (C6H1206) into 125 g of water. Ethylene glycol (C2H6O2) is a molecular compound that is used in many commercial anti-freezes. An aqueous solution of ethylene glycol is used in vehicle radiators to prevent the water in the radiator from freezing by lowering its freezing point. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500. g of water. Step 1: List the known quantities and plan the problem. Step 1: Kf (H2O) = -1.86 °C/m Known: mass of C2H6O2 = 400. g molar mass of C2H6O2 = 62.08 g/mol mass of H20 = 500. g = 0.500 kg Kf (H20) = -1.86 °C/m *Kf- is the constant freezing point coy nso ted Unknown: Tf of solution = ? °C *Tf- is the total freezing point This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression. Step 2: e er Step 2: Solve. 0 mal 400. g C2H6O2×1 mol C2H6O2÷62.08 g C2H6O2=6.44 mol C2H6O2 6.44 mol C2H602÷0.500 kg H2O=12.9 m C2H.O2 ATf = Kf xm = -1.86°C/m x 12.9 m = -24.0°C -24.0°C svio ents Tf %3D The normal freezing point of water is 0.0°C. Therefore, since the freezing point decreases by 24.0°C, the freezing point of the solution is -24.0°C. 2enE CHAVEE HER O 0f

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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D. Freezing Point Calculation Directions: Study the sample problem and do the task given. Sample problem with solution, Task to do Ethylene glycol (C2H6O2) is a molecular compound Calculate the freezing point of a solution that is used in many commercial anti-freezes. An aqueous solution of ethylene glycol is used in. vehicle radiators to prevent the water in the radiator from freezing by lowering its freezing point. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500.g of water. repared by dissolving 27.56 g of glucose (C6H1206) into 125 g of water. Step 1: List the known quantities and plan the problem. Step 1: Kf (H2O) = -1.86 °C/m %3D Known: mass of C2H6O2 = 400. g molar mass of C2H6O2 = 62.08 g/mol mass of H20 = 500. g = 0.500 kg Kf (H20) = -1.86 °C/m *Kf- is the constant freezing point %3D %3D %3D %3DD ni noiulge Unknown: Tf of solution = ? °C *Tf- is the total freezing point This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression. %3D Step 2: Step 2: Solve. 400. g C2H6O2 × 1 mol C2H6O2÷62.08 g C2H6O2=6.44 mol C2H6O2 6.44 mol C2H602÷0.500 kg H2O=12.9 m C2H6O2 ATf = Kf xm = -1.86°C/m x 12.9 m = -24.0°C Tf = -24.0°C cimal %3D The normal freezing point of water is 0.0°C. Therefore, since the freezing point decreases by 24.0°C, the freezing point of the solution is -24.0°C. O THE 3H TAO 3 ARORCTH 13
Transcribed Image Text:D. Freezing Point Calculation Directions: Study the sample problem and do the task given. Sample problem with solution, Task to do Ethylene glycol (C2H6O2) is a molecular compound Calculate the freezing point of a solution that is used in many commercial anti-freezes. An aqueous solution of ethylene glycol is used in. vehicle radiators to prevent the water in the radiator from freezing by lowering its freezing point. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500.g of water. repared by dissolving 27.56 g of glucose (C6H1206) into 125 g of water. Step 1: List the known quantities and plan the problem. Step 1: Kf (H2O) = -1.86 °C/m %3D Known: mass of C2H6O2 = 400. g molar mass of C2H6O2 = 62.08 g/mol mass of H20 = 500. g = 0.500 kg Kf (H20) = -1.86 °C/m *Kf- is the constant freezing point %3D %3D %3D %3DD ni noiulge Unknown: Tf of solution = ? °C *Tf- is the total freezing point This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression. %3D Step 2: Step 2: Solve. 400. g C2H6O2 × 1 mol C2H6O2÷62.08 g C2H6O2=6.44 mol C2H6O2 6.44 mol C2H602÷0.500 kg H2O=12.9 m C2H6O2 ATf = Kf xm = -1.86°C/m x 12.9 m = -24.0°C Tf = -24.0°C cimal %3D The normal freezing point of water is 0.0°C. Therefore, since the freezing point decreases by 24.0°C, the freezing point of the solution is -24.0°C. O THE 3H TAO 3 ARORCTH 13
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