You have a 4M aqueous solution of ethylene glycol (62 g / mol molar mass). You want to recover 49 ml of pure ethylene glycol from this solution. What is the minimum amount of aqueous solution required to obtain the desired ethylene glycol. density of ethylene glycol is 1.11 g / ml
You have a 4M aqueous solution of ethylene glycol (62 g / mol molar mass). You...
An solution of antifreeze is prepared by mixing 21.0mL of ethylene glycol (d = 1.11 g/mL; molar mass = 62.07 g/mol) with 50.0 mL H2O (d = 1.00 g/mL) at 25°C. If the density of the antifreeze solution is 1.07 g/mL, what is its molarity?
A solution of antifreeze is prepared by mixing 21.0mL of ethylene glycol (d = 1.11 g/mL; molar mass = 62.07 g/mol) with 50.0 mL H2O (d = 1.00 g/mL) at 25°C. If the density of the antifreeze solution is 1.07 g/mL, what is its molarity (M)?
An aqueous ethylene glycol (HOCH2CH2OH, FW=62.07 g/mol) solution with a mass of 220.7 mg is titrated with 45.3 mL of 0.0671 M Ce4 in 4 M HClO4. The solution is held at 60 °C for 15 min to oxidize the ethylene glycol to formic acid (HCO2H) and carbon dioxide. The excess Ce4+ is titrated with 11.73mL of 0.0449 M Fe2+ to a ferroin end point. What is the mass percent of ethylene glycol in the unknown solution? mass percent =__________%
An aqueous ethylene glycol (HOCH2CH2OH, FW = 62.07 g/mol) solution with a mass of 253.1 mg is titrated with 49.5 mL of 0.0895 M Ce4+ in 4 M HCIO4. The solution is held at 60°C for 15 minutes to oxidize the ethylene glycol to formic acid (HCO2H) and carbon dioxide. The excess Ce4+ is titrated with 10.81 mL of 0.0439 M Fe2+ to a ferroin end point. What is the mass percent of ethylene glycol in the unknown solution? Number
The liquid in automobile cooling systems is prepared by dissolving ethylene glycol (HOCH_2CH_2OH) in water. Ethylene glycol has a molar mass of 62. 07 g/mol and a density 1. 115 g/ml at 50. 0 degree C. Calculate the vapor pressure at 50 degree C of a coolant solution that is 51. 0: 49. 0 ethylene glycol to water by volume At 50. 0degree C the density of water is 0. 9880 g/ml, and its vapor pressure is 92 torr. The...
Ethylene glycol is used in automobile radiators as an antifreeze. Use the following information to determine the freezing point of an antifreeze solution made by mixing 2.5 L of ethylene glycol with 2.5 L of water. Ethylene glycol is essentially nonvolatile and it does not dissociate in water. Molar mass of ethylene glycol = 62.1 g/mol; density of ethylene glycol = 1.11 g/mL; density of water = 1.00 g/mL; Kf for water = 1.86 degrees Celsius kg/mol
A solution of antifreese is prepared by mixting 34.0mL of ethylene glycol (d=1.11g/mL; molar mass =62.07 g/mol) with 50.0 mL H20 (d=1.00 g/mol) at 25 degrees celsius. if the density of the antifreeze solution is 1.07 g/ mL, what is the molarity?
A 8.5 mass % aqueous solution of ethylene glycol (HOCH2CH2OH) has a density of 1.34 g/mL. Calculate the molarity of the solution. ANSWER: 1.84 +- 2%
An aqueous antifreeze solution is 31.0 % ethylene glycol (C2 H4 O2) by mass. The density of the solution is 1.05 g/cm3. Calculate the molality, molarity and mole fraction of the ethylene glycol Molality mol/kg Molarity mol/L Mole fraction
Antifreeze is an 50% by mass aqueous solution of the non-electrolyte ethylene glycol (C2H6O2). What is the freezing point of this solution? The Kf for water is 1.86oC∙kg∙mol−1 and assume a density of 1.00g/mL.