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Answers to Problems of the A Groups
Wt N that dissolves in 13.4 ml = (13.4 ml W) (
39 '*
8 N ) = 5.3 g N
\ 100 ml W/
Filter off 100 - 5.3 = 94.7 g pure N at 100.0°C
Step 3. Adjust water volume to that required to dissolve 5.3 g N at 0.0°C.
Vol W = (5.3 g N) I '°°
ml ™) = 14.9 ml W
\ 35.7 g N /
Wt K that dissolves in 14.9ml = (14.9 ml) (
4 '
9 8 K ] = Q.73 g K
\ 100 ml W/
Filter off 13.7 - 0.7 = 13.0 g pure K at 0.0°C
Step 4. Adjust water volume to that required to dissolve 0.73 g K at 100. 0°C.
Vol W = (0.73 g K) - ~ - = 0.72 ml W
\ 1UZ g K. /
/ 39 1 e NN
Wt N that dissolves in 0.80 ml = (0.80 ml W) I .„ ' ° „,) = 0.31 g N
\ 100 ml W/
Filter off 5.3 - 0.3 = 5.0 g pure N at 100. 0°C
After 2 cycles, you now have 86.3 + 13 = 99.3 g pure K 2 Cr 2 O 7
and 94.7 + 5 = 99.7 g pure NaCl
(c) Step 1. Treat the mixture with enough water (W) to dissolve 50 g KNO 3 (K) at
100.0°C.
Wt Ce 2 (SO 4 ) 3 (C) dissolved in 20.3 ml = (20.3 ml W) ( ?'*
e ,^ lf ) = 0.50 g C
\ 100 ml W/
Filter off 50.0 - 0.5 = 49.5 g pure Ce 2 (SO 4 ) 3 at 100°C
Step 2. Adjust water volume to that needed to dissolve 0.5 g C at 0.0°C.
Wt K dissolved in 5 ml = (5.0 ml W) ( ,'
3 '
3 g , ^} = 0.7 g K
\ 100 ml W/
Filter off 50.0 - 0.7 = 49.3 g pure KNO 3 at 0.0°C
There is no point in carrying this recrystallization any further.
(d) Step I. Treat the mixture with enough water (W) to dissolve 100 g NH 4 C1 (N) at
0.0°C.
/ 100 ml W\ =
I 29.7 gN/ ~
VolW = (100gN) '„ '"' " =336.7 ml W