Do sample problem 13.8 in the 8th ed silberberg book. you add 1.4 kg of ethylene glycol (c2h6o2) antifreeze to 4,192 g of water in your car's radiator. what is the boiling point of the solution? the kb for water is 0.512 °c/m. enter to 2 decimal places.

The solution boils at 102.76 °C

Explanation:

Δt = m* Kb x i  

i = 1 with 1 particle in solution

m = molality

Kb = 0.512 °C/molal

Step 2: Calculate moles C2H6O2

molar mass of  C2H6O2 = 62.068 g/mol

Calculate number of moles:

Moles = Mass / molar mass

Moles = 1400 grams / 62.068 g/mol

Moles C2H6O2 = 22.56 moles

Step 3: Calculate molality

these are in 4192 g of water:

22.56 moles / 4.192 kg water

⇒ moles / kg of water = 5.38 moles / Kg = m olal

Step 4: Calculate  ΔT

ΔT= Kb * molal = 5.38 molal* 0.512 °C/m

ΔT = 2.76 °C

Boiling point = 100°C + 2.75 °C = 102.76 °C

the solution boils at 102.76 °C

heat or electricity in the atmosphere- c.

pie chart because it is in percents ( % )