Calculate the freezing point of the following solutions, assuming complete disso

Question

Calculate the freezing point of the following solutions, assuming complete dissociation.

Part A

0.110m K2S tf =Degrees C

Part B

19.5g of CuCl2 in 410g water tf =Degrees C

Part C

5.2% NaNO3 by mass (in water) tf =Degrees C

Part D

Calculate the boiling point of the solution in part A, assuming complete dissociation. tb =Degrees C

Part E

Calculate the boiling point of the solution in part B, assuming complete dissociation. tb =Degrees C

Part F

Calculate the boiling point of the solution in part C, assuming complete dissociation. tb =Degrees C

Explanation / Answer

F.P depression constant of water = Kf = 1.86 0C/m ; B.P depression constant of water = 0.52 0C/m

B.P of pure water = 100 0C ; F.P of pure water = 0 0C

1) K2S will completely dissociate into 2 K+ ions and 1 S2- ion.Hence, the Vant Hoff's factor , i = 3

molality of solution = 0.11 m

Thus, depression in F.P = i*Kf*m = 3*1.86*0.11 = 0.6138 0C

Thus, F.P of the solution = F.P of pure water - depression in F.P = -0.6138 0C..........(a)

Elevation in B.P = i*Kb*m = 3*0.52*0.11 = 0.1716 0C

Thus, B.P of the solution = B.P of pure water - elevation in B.P = 100.1716 0C..............(b)

2) Molar mass of CuCl2 = 134.5 g/mole

Thus, moles of CuCl2 in 19.5 g of it = mass/molar mass = 19.5/134.5 = 0.145

mass of water in kg = 0.41

Ths, molality of the solution = moles of CuCl2/mass of water in kg = 0.145/0.41 = 0.354 m

Now,

CuCl2 will completely dissociate into 2 Cl- ions and 1 Cu2+ ion.Hence, the Vant Hoff's factor , i = 3

molality of solution = 0.354 m

Thus, depression in F.P = i*Kf*m = 3*1.86*0.354 = 1.97532 0C

Thus, F.P of the solution = F.P of pure water - depression in F.P = -1.97532 0C..........(a)

Elevation in B.P = i*Kb*m = 3*0.52*0.354 = 0.55224 0C

Thus, B.P of the solution = B.P of pure water - elevation in B.P = 100.55224 0C..............(b)

3) Molar mass of NaNO3 = 85 g/mole

Let there be 100 g of the solution

Thus, mass of NaNO3 = 5.2 g

Now, moles of NaNO3 = mass/molar mass = 5.2/85 = 0.0612

Mass of solvent = mass of solution - mass of NaNO3 = 94.8 g = 0.0948 Kg

Thus, molality of the solution = moles of NaNO3/mass of solvent in kg = 0.0612/0.0948 = 0.645

Now, NaNO3 will completely dissociate into 1 Na+ ions and 1 NO3- ion.Hence, the Vant Hoff's factor , i = 2

molality of solution = 0.645 m

Thus, depression in F.P = i*Kf*m = 2*1.86*0.645 = 2.401 0C

Thus, F.P of the solution = F.P of pure water - depression in F.P = -2.401 0C..........(a)

Elevation in B.P = i*Kb*m = 2*0.52*0.645 = 0.671 0C

Thus, B.P of the solution = B.P of pure water - elevation in B.P = 100.671 0C..............(b)

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