Identify the compound with the highest boiling point. Define triple point in a p
ID: 1009208 • Letter: I
Question
Identify the compound with the highest boiling point. Define triple point in a phase diagram. The temperature, pressure, and density for a gas. The temperature at which the boiling point equals the melting point. The temperature and pressure where liquid, solid, and gas arc equally stable and are in equilibrium. The temperature that is unique for a substance. The temperature at which the solid and liquid co-exist. Calculate the boiling point of a solution of 500.0 g of ethylene glycol (C_2H_6O_2) dissolved in 500.0 g of water. Kf = 1.86degree C/m and Kb = 0.512degree C/m. Use 100degree C as the boiling point of water. Choose the aqueous solution that has the highest boiling point. You should assume ideal van't Hoff factors where applicable. The first-order decomposition of N_2O at 1000 K has a rate constant of 0.76 s^-1. If the initial concentration of N_2O is 10.9 M, what is the concentration of N_2O after 9.6 s?Explanation / Answer
Ans 1- Among these,HF will have highest boiling point because among HF molecules, there will be greatest
hydrogen bonding present. So, HF molecules would be strongly bonded with hydrogen bond, hence its
boiling point would be high.
Ans 2 - Triple point is the temperature and pressure where liquid,solid and gas are equally stable and are in
equilibrium.
So, option C) is correct.
Ans 3 - Here, we would use elevation in boiling point formula.
Delta Tb = Kb .m
Here delta Tb = change in boiling point, Kb = boiling point constant , m = molality
delta Tb = Tb- T0b
Here, T0b = boiling piint of pure solvent, Tb = boiling point of solution
Given - Mass of solute(ethylene glycol) = 500 g , Molar mass of ethylene glycol = 12x2 + 1x 6 + 16x 2 = 62 g/mol
Mass of solvent(water) = 500 g =0.500 kg , Kb = 0.512 degree celsius/m
T0b = 100 degree celsius, Tb = ?
Now, first of all, we will find molality of solution.
Molality(m) = No. of moles of solute/Mass of solvent in Kg.
Moles of solute = Mass of solute/Molae mass of solute = 500/ 62
m = (500/62)/ 0.500
m = 16.1290 mol/Kg
Now, substituting value of T0b, Kb and m, we get value of Tb.
Tb - 100 = 0.512 x 16.1290
Tb - 100 = 8.2580
Tb = 100 + 8.2580
Tb = 108.2580 degree celsius