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. Exp. 3.00 Page 133 B REPORT FORM: Name 100.00 g of an aqueous solution contains 14.00 g of KCI. The density of the solution
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Answer #1

Molar mass of KCl,

MM = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

mass(KCl)= 14.00 g

use:

number of mol of KCl,

n = mass of KCl/molar mass of KCl

=(14 g)/(74.55 g/mol)

= 0.1878 mol

a)

Mass of solvent = mass of solution - mass of KCl

= 100.0 g - 14.00 g

= 86.00 g

= 0.08600 Kg

Use:

Molality = number of mol / mass of solvent in Kg

= 0.1878 mol / 0.08600 Kg

= 2.18 molal

Answer: 2.18

b)

Volume of solution = mass of solution / density of solution

= 100.0 g / 1.081 g/mL

= 92.51 mL

= 0.09251 L

use:

Molarity,

M = number of mol / volume in L

= 0.1878/0.09251

= 2.03 M

Answer: 2.03 M

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