Question

Ethylene glycol (C₂H₆O₂) is used as an additive to the water in your automobile to lower its freezing point. A solution of ethylene glycol in water has a freezing point of -4.10°C. How many grams of ethylene glycol must be added to 1000 g of this solution to lower the freezing point to -11.00 °C? (Kf for water is 1.86°C･kg/mol).

Answer 1

for depression in freezing point (colligative property) We have! AT =Kfm Molality of solution Constant for given solvent changes in freezing point. Here, solvent is water and its normal freezing point is 0°C. So, we can find out the molality when temp art freezing point is ~4.10°C 8 -11.00°. this can be calculated as: =) 4.10% = 1086 °C kg/mol X m, m, = 2.20 mol/ieg - 11.00% = 1.86 °C kg/mol Xm2 m₂ = 5.91 mol/kg * do 1000g of solution with molality m, -2.20 molsko mass of solvent is about 880g. molecular mass of ethylene glycol is 62 m, = 2.20 mollag eans 136,4g of ethylene glycol in 100og H₂O Let the amount of solvent be x. 13 6.4 tk = 1000g x = 879.97 g *moles of ethylene glycol to lower F.P. to -11°C 5. Il mol 10 x 879.97 g 5.2 mol. 1000g * Intially present mole = 136 42 136.4X580 - 1.94 nol Tvoox62 1000x62 * Moles required to be added = 5.20-1.9423.26. * Mass of ethylene glycol – 3.26x62 = 120 T000

for depression in freezing point (colligative property) We have! AT =Kfm Molality of solution Constant for given solvent changes in freezing point. Here, solvent is water and its normal freezing point is 0°C. So, we can find out the molality when temp art freezing point is ~4.10°C 8 -11.00°. this can be calculated as: =) 4.10% = 1086 °C kg/mol X m, m, = 2.20 mol/ieg - 11.00% = 1.86 °C kg/mol Xm2 m₂ = 5.91 mol/kg * do 1000g of solution with molality m, -2.20 molsko mass of solvent is about 880g. molecular mass of ethylene glycol is 62 m, = 2.20 mollag eans 136,4g of ethylene glycol in 100og H₂O Let the amount of solvent be x. 13 6.4 tk = 1000g x = 879.97 g *moles of ethylene glycol to lower F.P. to -11°C 5. Il mol 10 x 879.97 g 5.2 mol. 1000g * Intially present mole = 136 42 136.4X580 - 1.94 nol Tvoox62 1000x62 * Moles required to be added = 5.20-1.9423.26. * Mass of ethylene glycol – 3.26x62 = 120 T000