CH. 12 HW Solutions . Include the formula Steps involved in an orderly fashion C

Question

CH. 12 HW Solutions . Include the formula Steps involved in an orderly fashion Calculate the freezing point of a solution containing 5.0 grams of KCI in 550.0 grams of water. The molal-freezing-point-depression constant (Kf) for water is 1.86 /m. 1. A)-045CB)+0.450C (C)423 0C :D)+0.23 0C E)1250C x9 033 An aqueous solution has a normal boiling point of 103.0 what is the freezing point of this solution? For water Kb = 0.51 in and Kf-1.86 m. A)-0, 82 B)-3.0·3.6°C D)-11°C 2. What is the expected freezing point of a 0.50 m solution of Li2S04 in water? Kf for water is 1,86°C/m. A)-0.93 3. B)-1.9°CC),2.8°C D)-65°C 0.00°C-0.93"c=-0.93 At 20°C, an aqueous solution that is 24.0% by mass in ammonium chloride has a density of 1.0674 g/mL. What is the molarity of ammonium chloride in the solution? The formula weight of NH4CI is 53.50 g/mol. 4, A)5.90 B)0.479C)4.79D)0.0445 E)22. Volume of Saluhont

Explanation / Answer

1.

Mass of KCl = 5.0 g.

Molar mass of KCl = 39.0 + 35.5 = 74.5 g/mol

Moles of KCl = mass / molar mass = 5.0 / 74.5 = 0.0671 mol

Mass of solvent = 550.0 g. = 0.550 kg.

Molality = moles of solute / mass of solvent in kg

m = 0.0671 / 0.550

m = 0.122 m

Kf of water = 1.86 0C/m

Depression in freezing point = Kf * m

0.00 - Tf = 1.86 * 0.122

Tf = - 0.227 0C = - 0.23 0C

(C)

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