What is the freezing point of a nonionizing antifreeze solution containing 388g ethylene glycol $C_{2} H_{6} O_{2}$ $C_2H_6O_2$ and 409 g of water?

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Answer 1 · 2018-01-04T17:54:08

The freezing point = $- {28.42}^{o} C$ $- 28.42^oC$

$∆T_b$ = $i*K_f*m$

mass of ethylene = $388 g$

molar mass = $62.07$ $g$ / $mol$

no. of moles = $388$ / $62.07$ = $6.25$ moles

mass of solvent = $409 g$ = $0.409 kg$

molality = $6.25$ moles/ $0.409 kg$ = $15.28 m$

Therefore,

$∆T_f = i*K_f*m$
$∆T_f = 1.86^oC/m xx 15.28 m$

$∆T_f = 28.42^o C$

$T_f = 0^oC - 28.42^ oC = -28.42^ oC$

What is the freezing point of a nonionizing antifreeze solution containing 388g ethylene glycol C_2H_6O_2C2H6O2C_2H_6O_2 and 409 g of water?