Answers to Problems of the A Groups
461
Wt KClO 3 (K) that dissolves in 336.7 ml = (336.7 ml W) ( ,^'
8 g ,^ lf ) = 12.8 g K
\ 100 ml W/
Filter off 50.0 - 12.8 = 37.2 g pure K at 0.0°C
Step 2. Adjust water volume to that needed to dissolve 12.8 g K at 100. 0°C.
g
Wt N that dissolves in 22 ml = (22 ml W) (
?5 '
8 g N ) = 16.7 g N
\ 100 ml W/
Filter off 100.0 - 16.7 = 83.3 g pure N at 100. 0°C
Step 3. Adjust water volume to that needed to dissolve 16.7 g N at 0.0°C.
Vol W = (16.7 g N )
= 56.2 ml W
Wt K that dissolves in 56.2 ml = (56.2 ml W) ( "* '
8 8
,
K } = 2. 1 g K
\ 100 ml W/
Filter off 12.8 - 2.1 = 10.7 g pure K at 0.0°C
Step 4. Adjust water volume to that needed to dissolve 2.1 g K at 100. 0°C.
O. 1 g Is.
Wt N that dissolves in 4 ml = (4.0 ml W) (
7 f'
8 ^ ^,) = 3.0 g N
\ 100 ml W/
Filter off 16.7 - 3.0 = 13.7 g pure N at 100. 0°C
After 2 cycles, you have 83.3 + 13.7 = 97.0 g pure NH 4 C1
and 37.2 + 10.7 = 47.9 g pure KC1O 3