Answer:
rise,cool..................
Answer:
P' = 41.4 mmHg → Vapor pressure of solution
Explanation:
ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')
Xm = Mole fraction for solute (Moles of solvent /Total moles)
Firstly we determine the mole fraction of solute.
Moles of solute → Mass . 1 mol / molar mass
20.2 g . 1 mol / 342 g = 0.0590 mol
Moles of solvent → Mass . 1mol / molar mass
60.5 g . 1 mol/ 18 g = 3.36 mol
Total moles = 3.36 mol + 0.0590 mol = 3.419 moles
Xm = 0.0590 mol / 3.419 moles → 0.0172
Let's replace the data in the formula
42.2 mmHg - P' = 42.2 mmHg . 0.0172
P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)
P' = 41.4 mmHg
Answer:
Molality = 8.57 m
Explanation:
Given data:
Molarity of solution = 5.73 M
density = 0.9327 g/mL
Molality of solution = ?
Solution:
Molality = moles of solute / kg of solvent.
Kg of solvent:
Mass of 1 L solution = density× volume
Mass of 1 L solution = 0.9327 g/mL × 1000 mL
Mass of 1 L solution = 932.7 g
Mass of solute:
Mass of 1 L = number of moles × molar mass
Mass = 5.73 mol × 46.068 g/mol
Mass = 263.97 g
Mass of solvent:
Mass of solvent = mass of solution - mass of solute
Mass of solvent = 932.7 g - 263.97 g
Mass of solvent = 668.73 g
In Kg = 668.73 /1000 = 0.6687 Kg
Molality:
Molality = number of moles of solute / mass of solvent in Kg
Molality = 5.73 mol / 0.6687 Kg
Molality = 8.57 m
To reduce a haloalkane, it has to be eliminated first. This is because 2 bromopropanes are a saturated compound that can not be reduced by a saturated compound. Dehydrohalogenation of the haloalkanes into the form of propene and hydrogen bromide can lead to the elimination. The propene is then reduced to propane.
