If an aqueous solution freezes at -5.58 Celsius, at what temperature will it boil?

Question

thanks!

Answer 1

delta T = kf m

5.58 = 1.86 m

m = 3

to get the boiling point we use the equation

delta T = kb m

kb = 0.512

delta T = 0.512 x 3 = 1.54 °C

the original b.p is 100 °C ( corresponding pure water ) ; it will be 101.54 °C

Answer 2

the depression of freezing point is -5.58°C
Knowing that 1 osmole depreses of 1.86°C , the solution contains -5.58/-1.86 = 3 Osmole
the constant ebullioscopic of water is 0.51° so the elavation of boiling point is 0.51*3 =1.53°C

The boiling point is 101.53°C

Answer 3

delta T = kf m

5.58 = 1.86 m

m = 3

to get the boiling point we use the equation

delta T = kb m

kb = 0.512

delta T = 0.512 x 3 = 1.54 °C

the original b.p is 100 °C ( corresponding pure water ) ; it will be 101.54 °C

Answer 4

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