iPad 11:22 PM @ 18%. ], ufl.instructure.com D Question 2 1 pts What is the molal

Question

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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