What is the freezing point (in C) of each of the solutions below? For water, K f

Question

What is the freezing point (in C) of each of the solutions below? For water, Kf = 1.86(C kg)/mol. The vapor pressure of water at 45.0C is 71.93mm Hg.

Part A

A solution of 14.5 g of urea, CH4N2O, in 164 g of water at 45.0C.

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Part B

A solution of 13.0 g of LiCl in 164 g of water at 45.0C, assuming complete dissociation.

What is the freezing point (in C) of each of the solutions below? For water, Kf = 1.86(C kg)/mol. The vapor pressure of water at 45.0C is 71.93mm Hg.

Part A

A solution of 14.5 g of urea, CH4N2O, in 164 g of water at 45.0C.

mp =   C  

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Part B

A solution of 13.0 g of LiCl in 164 g of water at 45.0C, assuming complete dissociation.

mp =   C  

Explanation / Answer

Part A:

We know that T f = Kf x m

Where

T f = depression in freezing point

= freezing point of pure solvent – freezing point of solution

= 0-Tf

= -Tf

K f = depression in freezing constant=1.86 oC// m

m = molality of the solution

= ( mass / Molar mass ) / weight of the solvent in Kg

= (14..5g/60(g/mol))/0.164 kg

= 1.473 m

Plug the values we get Tf=-2.74 oC

Part B:

We know that T f = Kf x m

Where

T f = depression in freezing point

= freezing point of pure solvent – freezing point of solution

= 0-Tf

= -Tf

K f = depression in freezing constant= 1.86oC/m

m = molality of the solution

= ( mass / Molar mass ) / weight of the solvent in Kg

= (13..6. g/42.4(g/mol))/0.164 kg

= 1.96 m

Plug the values we get Tf= -3.64 oC

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