To determine which of the given compounds will give a silver mirror with ammoniacal silver nitrate, we need to understand the chemistry behind the Tollens' test. This test is used to identify aldehydes. When aldehydes react with Tollens' reagent, which is a solution of ammoniacal silver nitrate, they are oxidized to carboxylic acids, and the reagent forms a silver mirror on the test tube as metallic silver is precipitated. Let's analyze each compound given:
Based on the analysis above, the correct compounds that will give a silver mirror with ammoniacal silver nitrate are Formic acid (A), Formaldehyde (B), and Benzaldehyde (C). Therefore, the correct answer is:
A, B, and C only
To determine which compounds will give a silver mirror with ammoniacal silver nitrate, we need to consider the Tollens' test. This test is generally used to identify aldehydes. The Tollens' reagent, which contains ammoniacal silver nitrate, oxidizes aldehydes to carboxylic acids, and in the process, the silver ion is reduced to metallic silver, forming a silver mirror on the inner walls of the test tube.
Therefore, the compounds that will give a silver mirror with ammoniacal silver nitrate are Formic acid, Formaldehyde, and Benzaldehyde. Thus, the correct answer is:
A, B, and C only
(i) Explain Aldol condensation with example.
(ii) How are the following conversions achieved:
(a) Benzene Benzaldehyde, (b) Ethanoic acid ethanol.
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]