iPad 11:22 PM @ 18%. ], ufl.instructure.com D Question 2 1 pts What is the molal
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iPad 11:22 PM @ 18%. ], ufl.instructure.com D Question 2 1 pts What is the molality of a 32.925 wt% solution of CaCl2 in water? Assume the density of the solution is equal to that of pure water (1.000 g/mL). Express your response to three digits after the decimal. DQuestion 3 1 pts What is the theoretical boiling point of a 36.833 wt% solution of CaCl2 in water? Kb(H2O-0.52 "C/m. Tb?_ 100.00 ? Assume the density of the solution is equal to that of pure water (1.000 g/mL). Express your response to three digits after the decimal.Explanation / Answer
2)
Let mass of solution be 1 Kg = 1000 g
mass of CaCl2 = 32.925 % of mass of solution
= 32.925*1000.0/100
= 329.25 g
mass of solvent = mass of solution - mass of solute
mass of solvent = 1000 g - 329.25 g
mass of solvent = 670.75 g
mass of solvent = 0.67075 Kg
Molar mass of CaCl2,
MM = 1*MM(Ca) + 2*MM(Cl)
= 1*40.08 + 2*35.45
= 110.98 g/mol
mass(CaCl2)= 329.25 g
use:
number of mol of CaCl2,
n = mass of CaCl2/molar mass of CaCl2
=(329.25 g)/(1.11*10^2 g/mol)
= 2.967 mol
m(solvent)= 0.67075 Kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(2.967 mol)/(0.67075 Kg)
= 4.423 molal
Answer: 4.423 molal
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