The steam volatile compounds among the following are:
Choose the correct answer from the options given below:
Steam volatile compounds are typically those that are capable of being vaporized or evaporated easily at relatively low temperatures. Generally, compounds with hydrogen bonding or low molecular weight are steam volatile.
Among the options:
- (A) \( {C}_6{H}_4{OH}{NO}_2 \): This is a nitrophenol compound, which is steam volatile due to the phenolic group that can form hydrogen bonds and is low molecular weight.
- (B) \( {C}_6{H}_4{NH}_2{NO}_2 \): This is a nitroaniline compound, which can also be steam volatile due to the amine group that can form hydrogen bonds.
- (C) \( {C}_6{H}_4{OH}{NH}_2 \): This is an amphenol compound, and although it contains an amine and phenolic group, it has a higher molecular weight and is less likely to be steam volatile.
- (D) \( {C}_6{H}_5{OH} \): This is phenol, which is steam volatile but not included in the right matching with other steam volatile candidates. Thus, the correct answer is (1).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,