The question asks whether the Kjeldahl method is applicable for estimating nitrogen in pyridine and whether the nitrogen in pyridine can be easily converted to ammonium sulphate via this method.
This is false. The nitrogen in pyridine is part of a stable aromatic ring, which does not get converted into ammonia easily in the Kjeldahl digestion process. Hence, the Kjeldahl method is not appropriate for determining nitrogen in pyridine.
This is also false. As mentioned earlier, the nitrogen in pyridine is not readily convertible to ammonium sulphate because of its stable aromatic nature.
Conclusion: Both Statement I and Statement II are false. The correct answer is: Both Statement I and Statement II is false.
The given question involves understanding the suitability of the Kjeldahl method for estimating nitrogen content in different compounds, particularly pyridine.
To determine the correctness of these statements, let's delve into the principles of the Kjeldahl method:
The Kjeldahl method is a widely used procedure for estimating the nitrogen content in organic compounds. It is particularly effective for compounds containing nitrogen in the form of amines, amides, imides, and heterocyclic compounds with readily convertible nitrogen atoms. However, it may not be suitable for certain nitrogen-containing compounds.
Analysis of the Statements:
Conclusion:
Both statements are false because the Kjeldahl method is not suitable for estimating nitrogen in pyridine due to the reasons outlined above.
Thus, the correct answer is: Both Statement I and Statement II is false.
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
