Question

I need help in find the solution and understanding the work process behind solve the problems step by step. I would appreciate the help with these problems

1. What is the treezing point of a solution prepared by dissolving 20 g of NaCI (MW 58.4 g mol) in 150 ml. of water? The freezing point depression constant, K, for water 53.4 2. What is the osmotic pressure formed by dissolving 44 2 mg of aspirin (C,l1,0) in 0.358 L of water at 25 "C? Calculate the molality of a solution made by dissolving 5.25 g of glucose (C,HO) in 30 mL of water at 25 C 3.

Answer 1

1.

Moles of NaCl = mass / molar mass = 20. / 58.4 = 0.342 mol

Mass of solvent = 150 g. = 0.150 kg.

molality = moles of solute / mass of solvent in kg

m = 0.342 / 0.150

m = 2.28 m

depression in freezing point = Kf * m

0.00 - T_f = 1.86 * 2.28

T_f = Freezing point of solution = - 4.25 0C

2.

moles of aspirin = mass / molar mass = 0.0442 / 180.16 = 0.000245 mol

VOlume of solution = 0.358 L

Molarity of solution = moles of solute / volume of solution = 0.000245 / 0.358 = 0.000685 M

Therefore,

Osmotic pressure = C R T = 0.000685 * 0.0821 * 298.15 = 0.0168 atm

3.

moles of glucose = mass / molar mass = 5.25 / 180. = 0.0292 mol

Mass of solvent = density * volume = 1 * 30 = 30 g. = 0.030 kg.

Molality = moles of glucose / mass of solvent in kg = 0.0292 / 0.030 = 0.972 m