Answers to Problems of the A Groups
459
Vol W = (74.6 gK)=267.4 ml W
Wt NH 4 C1O 4 (S) dissolved in 267.4 ml = (267.4 ml W) ( |"'
7 g S } = 28.6 g S
\ 100 ml W/
Filter off 117.5 - 28.6 = 88.9 g pure S at 0.0°C
Step 2. Adjust water volume to that required to dissolve 28.6 g S at 100. 0°C.
Vol W = (28.6 g S) f
10
°
m ' f) = 60.5 ml W
V 47.3 g S /
Wt K dissolved in 60.5 ml = (60.5 ml W) ( ^ '
7 g
, ^\ = 34.3 g K
\ 100 ml W/
Filter off 74.6 - 34.3 = 40.3 g pure K at 100. 0°C
Step 3. Adjust water volume to that required to dissolve 34.3 g K at 0.0°C.
Wt S dissolved in 123 ml = (123 ml W) ( -^-f^l = 13.2 g S
V 100 ml W/
Filter off 28.6 - 13.2 = 15.4 g pure S at 0.0°C
Step 4. Adjust water volume to that required to dissolve 13.2 g S at 100. 0°C.
Vol W = (13.2 g S ) -
=27.9 ml W
V 47.3 g S /
Wt K dissolved in 27.9 ml W = (27.9 ml W) ( ^
S , ^\ = 15.8 g K
\ 100 ml W/
Filter off 34.3 - 15.8 = 18.5 g pure K at 100.0°C
After 2 cycles, you have 88.9 + 15.4 = 104.3 g pure NH 4 C1O 4 (S)
and 40.3 + 18.5 = 58.8 g pure KC1 (K).
(b) Step I, Treat the mixture with enough water (W) to dissolve 100 g NaCl (N) at
0.0°C.
Vol W = (100 g N) ( *°"
m ' T) = 280 ml W
\ 35.7 g N /
/ 4 9 e K \
Wt K 2 Cr 2 O 7 (K) dissolved in 280 ml = (280 ml W) ( .
s
I = 13.7 g K
Filter off 100 - 13.7 = 86.3 g pure K at 0.0°C
Step 2. Adjust the water volume to that needed to dissolve 13.7 g K at 100. 0°C.
Vol W = (13.7 gK)
= 13.4 ml W