- ARed solution
- ✓Blue solution
- CWhite precipitate
- DYellow colouration
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In the above reaction, $3.9\, g$ of benzene on nitration gives $4.92\, g$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is............. $\%$. (Round off to the Nearest Integer).
(Given atomic mass: $C : 12.0\, u , H : 1.0\, u$$O : 16.0\, u , N : 14.0\, u )$
Precipitate $X$ is
$(A)$ $Fe _4\left[ Fe ( CN )_6\right]_3$
$(B)$ $Fe \left[ Fe ( CN )_6\right]$
$(C)$ $K _2 Fe \left[ Fe ( CN )_6\right]$
$(D)$ $KFe \left[ Fe ( CN )_6\right]$
Among the following, the brown ring is due to the formation of
$(A)$ $\left[ Fe ( NO )_2\left( SO _4\right)_2\right]^{2-}$
$(B)$ $\left[ Fe ( NO )_2\left( H _2 O \right)_4\right]^{3+}$
$(C)$ $\left[ Fe ( NO )_4\left( SO _4\right)_2\right]$
$(D)$ $\left[ Fe ( NO )\left( H _2 O \right)_5\right]^{2+}$
$[Figure]$ $\xrightarrow[{2.\,C{H_3}I\,(l.\,eq.)}]{{1.\,{K_2}C{O_3}}}$
