Chem Mid Term Answer Key

Content type

User Generated

Subject

Chemistry

School

Tennessee State University

Course

chem 121

Type

Homework

Uploaded By

qbhtc758

Pages

1

Rating

1

C

10

A

19

C

2

B

11

B

20

C

3

B

12

A

21

B

4

D

13

C

22

D

5

C

14

C

23

C

6

D

15

E?

24

D

7

D

16

D

25

C

8

B

17

B

9

A

18

a

I think 15 is “E” but I can’t read the mass of CuSO4 isolated. It looks like 1.3 something making the %

lost (1 – ((2.369 – 1.3xx)/7.384)) * 100% ≈ 37.7%

Bonus

There is a problem with the bonus question!!!!

The correct equation to use is

dTfp = Kf ∗ m ∗ i

Where

Dtfp = change in freezing point = 2.82°C

Kf = cryoscopic constant = 1.86°C/m

m = molality = moles solute / kg solvent

i = van’t hoff factor = # ions 1 particle of solute dissociates into in solution = 2 for NaCl

why is the van’t hoff factor = 2 for NaCl in H2O?

1 NaCl → 1 Na(+) + 1 Cl(-)

That is…

1 particle → 2 ions

Therefore I = 2

Logic.

(1) Determine m

(2) Convert m and kg solvent to moles solute

(3) Convert moles solute to g salt

Solution

m =

dTfp

Kf ∗ i

=

2.82°C

1.86°

C

m

∗ 2

= 0.7581m

g salt =

3.0 kg ice

1

×

0.7581 mol NaCl

kg ice

×

58.4428g NaCl

mol NaCl

= 130g NaCl (2 sig figs)

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1 2 3 4 5 6 7 8 9 C B B D C D D B A 10 11 12 13 14 15 16 17 18 A B A C C E? D B a 19 20 21 22 23 24 25 C C B D C D C I think 15 is “E” but I can’t read the mass of CuSO4 isolated. It looks like 1.3 something making the % lost (1 – ((2.369 – 1.3xx)/7.384)) * 100% ≈ 37.7% Bonus Ther ...
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