As a research chemist, you are interested in studying the extent snd types of in

Question

As a research chemist, you are interested in studying the extent snd types of interactions in aqueous salt solutions. As part of this study, you weigh three samples of NaCl and dissolve each in 1.000 kg H2O. You then measure the freezing temperature of each solution and compare these temperatures to the freezing point of water. The data you collect are tabulated below. Explain the oberved results. Predict and briefly explain the result you would expect for a solution made up of 29.22g NaCl dissolved in 1.000 kg H2O.

Can you please help me!? I don't understand this at all! Please explain the steps. (especially the calculations)

g NaCl per 1.000 kg H2O deltaT fr *C (Freezing point Celsius) 5.845 0.348 0.585 0.0360 0.293 0.0182

Explanation / Answer

we know that

when a solute is added to the solvent

the freezing point of the solven decreases

this is called depression in freezing point

it is given by

dTf = i x Kf x m

here

i = vanthoff factor

Kf = molal freezing point depression constant of water = 1.86

m = molality

1) now consider 5.845 g of NaCl

we know that

moles = mass / molar mass

molar mass of NaCl = 58.44 g/mol

so

moles of NaCl = 5.845 / 58.44

moles of NaCl = 0.1

now

molality = moles of NaCl / mass of water (kg)

so

molality = 0.1 / 1 = 0.1

now

NaCl ---> Na+ + Cl-

we can see that

there are 2 particles after dissociation

so i = 2

now

dTf = 2 x 1.86 x 0.1

dTf = 0.372

we know that freezing point of water is 0 C

so

0 -Tf = 0.372

Tf = -0.372

so

freezing point of the 5.845 g NaCl solution is -0.372 C

2)

But

the observed result for dTf is 0.348 , this is because NaCl is not dissociated 100 %

now consider

dTf = i x Kf x m

using given values

0.348 = i x 1.86 x 0.1

i = 1.871

similarly for 0.585 g NaCl

moles of NaCl = 0.585 / 58.44 = 0.01

molality = 0.01 / 1 = 0.01

dTf = ix Kf x m

0.036 = i x 1.86 x 0.01

i = 1.9355

similarly for 0.293 g NaCl

moles of NaCl = 0.293 / 58.44 = 5 x 10-3

molality = 5 x 10-3 / 1 = 5 x 10-3

dTf = i x Kf x m

0.0182 = i x 1.86 x 5 x 10-3

i= 1.9516

now

consider the average for three value

i = ( 1.871 + 1.9355 + 1.9516) / 3

i = 1.9194

now

use this value for 29.22 g NaCl

moles of NaCl = 29.22 / 58.44 = 0.5

molality = 0.5 / 1 = 0.5

now

dTf = i x Kf x m

dTf = 1.9194 x 1.86 x 0.5

dTf = 1.785

so

the predicted value of dTF for 29.22 g NaCl is 1.785

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