A common antifreeze for car radiators is ethylene glycol, CH2(OH)CH2(OH). How ma

Question

A common antifreeze for car radiators is ethylene glycol, CH2(OH)CH2(OH). How many milliliters of this substance would you add to 5.9 L of water in the radiator if the coldest day in winter is -21°c? (The density and boiling point of ethylene glycol are 1.11 g-cm-3 and 470. K, respectively. Assume the density of water at its freezing point is 0.99987 g cm-3,and Kf for water is 1.86 K kg mol-1) mL Would you keep this substance in the radiator in the summer to prevent the water from boiling? Since the boiling point of ethylene glycol is Selectthe boiling point of water, adding it to the water in a radiator willSelectthe boiling point. Supporting Materials Periodic Table Constants and Factors Supplemental Data Submit Answer Save Progress Practice Another Version

Explanation / Answer

DTf = i*kf*m

DTf = T0-Tf = 0-(-21)

DTf = i*Kf*m

i = vanthoff factor of glycol = 1

Kf = frreezing point constant of water = 1.86 C/m

molality(m) = (w/M)*(1000/Wt)

w = wt of ethyleneglycol = ?

M = molarmass of ethyleneglycol = 62 g/mol

Wt = wt of water = 5.9*0.99987*10^3 = 5899.23 g

molality(m)= (x/62)*(1000/5899.23)

(0-(-21)) = 1*1.86*(x/62)*(1000/5899.23)

x = wt of ethyleneglycol =4129.46 grams

volume of ethyleneglycol = mass/density = 4129.46/1.11 = 3720.23 ml

yes, we can use.

greater than , increase

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