What is the normal boiling point in C of an antifreeze solution prepared by diss

Question

What is the normal boiling point in C of an antifreeze solution prepared by dissolving 617.4 g of ethylene glycol (C2H6O2) in 500.0 g of water? The molal boiling-point-elevation constant for water is given in the following table.

Express your answer to four significant figures and include the appropriate units.

KCkg)/mo KC.kg)/mol] Substance Benzene (C&Hs; Camphor (CioHieo Chloroform (CHCI) Diethyl ether (C,Hoo) Ethyl alcohol (C2H,O) Water (H,0) 2.64 5.95 3.63 2.02 1.22 0.51 5.07 37.8 4.70 1.79 1.99 1.86

Explanation / Answer

Lets calculate molality first

Molar mass of C2H6O2 = 2*MM(C) + 6*MM(H) + 2*MM(O)

= 2*12.01 + 6*1.008 + 2*16.0

= 62.068 g/mol

mass of C2H6O2 = 617.4 g

we have below equation to be used:

number of mol of C2H6O2,

n = mass of C2H6O2/molar mass of C2H6O2

=(617.4 g)/(62.068 g/mol)

= 9.947 mol

mass of solvent = 500 g

= 0.5 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(9.947 mol)/(0.5 Kg)

= 19.89 molal

lets now calculate deltaTb

deltaTb = Kb*m

= 0.51*19.8943

= 10.1461 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 10.1461

= 110.1461 oC

Answer: 110.1 oC

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know

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