Perhitungn.docx

Lampiran 2. Perhitungan

1. Penentuan Densitas Larutan (d) W - We

di =

d1 =

Wo - We 26,60 g

d2 =

25 g 26,49 g 25 g

x do x 0,995340 g/cm3 = 1,0590 g/cm3

x 0,995340 g/cm3 = 1,0546 g/cm3

25,76 g d3 = 25 g x 0,995340 g/cm3 = 1,0255 g/cm3

d4 =

25,38 g 25 g

x 0,995340 g/cm3 = 1,0104 g/cm3

25,18 g d5 = 25 g x 0,995340 g/cm3 = 1,0025 g/cm3

2. Penentuan Molalitas Larutan (m) m = m1 =

m2 =

m3 =

m4 =

1 d/M – BM/1000 1 = 3,3955 mmol/g 3 1,0590 g/cm - 58,5 g/mol 1000 3 mmol/cm3 1 1,0546 g/cm3 1,5 mmol/cm3

-

1 3 1,0255 g/cm 0,75 mmol/cm3

58,5 g/mol 1000

-

= 1,5514 mmol/g

58,5 g/mol 1000

= 0,7640 mmol/g

1 = 0,3793 mmol/g 1,0104 g/cm3 - 58,5g/mol 1000 0,375 mmol/cm3

m5 =

1 3 1,0025 g/cm

0,1875 mmol/cm3

-

58,5 g/mol 1000

= 0,1891 mmol/g

3. Penentuan Volume Molal Parsial () 1  = d (BM

1=

-

1000 W-Wo × Wo-We) m

1 1000 1,60 g -1 58,5 g . mol × ( ) 1,0590 g.cm-3 3,3955 mmol.g-1 25 g

= 37,4424 cm3/mol 2 =

1 1000 1,49 g (58,5 g.mol-1 × 25 g ) -3 -1 1,0546 g.cm 1,5514 mmol.g

= 19,0433 cm3/mol 3 =

1 1000 0,76 g (58,5 g.mol-1 × 25 g ) -3 -1 1,0255 g.cm 0,7640 mmol.g

= 18,2441 cm3/mol 4 =

1 1000 0,38 g (58,5 g.mol-1 × ) 1,0104 g.cm-3 0,3793 mmol.g-1 25 g

= 18,2365 cm3/mol 5=

1 1000 0,18 g (58,5 g.mol-1 × ) 1,0025 g.cm-3 0,1891 mmol.g-1 25 g

=18,2161 cm3/mol

4. Perhitungan  Regresi y = ax + b a. untuk x = 1,8426 y = 13,03 (1,8426) + 9,164 = 33,1730 b. untuk x = 1,2455 y = 13,03 (1,2455) + 9,164 = 25,3928 c. untuk x = 0,8740

y = 13,03 (0,8740) + 9,164 = 20,5522 d. untuk x = 0,6157 y = 13,03 (0,6157) + 9,164 = 17,1865 e. untuk x = 0,4348 y = 13,03 (0,4348) + 9,164 = 14,8294