part a part b part c part d If is very simple to determine how much the boiling

Question

part a

part b

part c

part d

If is very simple to determine how much the boiling point will increase or the freezing point will decrease for any solvent, if the number of particles of solute are known. Each solvent has two characteristic values (that can be looked up): the boiling point elevation constant, KBp, and the freezing point depression constant, KFp. These constants are multiplied by the molality, m, and the number of dissolved particles, i, to obtain the change in temperature, AT. It is important to note that this is AT from the original temperature, i is called the van't Hoff factor. water? If Tis calculated and the boiling point of pure water is 100°C, should the new boiling point with a solute be higher, lower or the same as that of pure Choose all of the correct answers that apply Higher - AT Lower

Explanation / Answer

part A )

By additing a solute the boiling temperature of solvent increases, which is measured as a colligative property called elevation in boiling point.

Thus

TBP(soln) = TBP(pure) + delta T

and the boiling point is HIGHER than water.

options 1 and 2

Part B)

The freezing temperature of a solution is less than that of pure solvent , and sis measured as depression in freezing point, a colligative property.

Thus TFP (sol) = TFP(pure) - delta T

and freezing point is lowered.

Thus options 1 and 3 are correct.

Part C)

The colligative property of elevation is boiling point is calculated using the expression

delta T (B) = i x m x K(fP)

where i = van't Hoff factor

m = molal concentration of solute

Kfp = freezing point constant for the solvent

Part D )

For NaCl which is an electrolyte assuming complete dissociation

NaCl -----------> Na+ + Cl-

thus van't Hoff factor of NaCl i =2

and molality = 0.13 m

Kfp = 1.86C/m

Thus delta T = 2x 0.13 mx 1.86 C/m

= 0.4836 C

Thus the final freezing point temperature of NaCL solution = 0C -0.4836 C

= -0.4836 C

= 272.6664 K

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