Analytical Chemistry Question 274
Question: The weight of a residue obtained by heating 2.76 g of silver carbonate is
[Pb. PMT 2004]
Options:
A) 2.76 g
B) 2.98 g
C) 2.16 g
D) 2.44 g
Show Answer
Answer:
Correct Answer: C
Solution:
$ 2Ag_2CO_3 $
$ \xrightarrow{\Delta } $
$ 4Ag+2CO_2+O_2 $ [(2 × 108) + 12 + 48] 4 × 108 2(216 + 12 + 48) 4 × 108 2 × 276 = 552 4 × 108 $ \because $ 552 gm silver carbonate gives silver = 432 gm. \ 2.76 gm silver carbonate gives $ \frac{432\times 2.76}{552}2.16gm $
BETA
