Aldehyde ketones & carboxylic Acid MCQs for NEET — Chemistry Questions with Answers

Orchids are known for having seeds that do not contain stored food. Unlike many seeds, orchid seeds are very small and lack endosperm, which is the tissue that provides nutrients to the developing plant embryo. This is why orchid seeds are often dependent on symbiotic relationships with fungi to germinate and grow.

In organic chemistry, C-C bond lengths depend on the type of bond between the carbon atoms. In $C_2H_2$ (acetylene), the carbon atoms are connected by a triple bond, which is shorter than double bonds (found in $C_2H_4$) and single bonds (found in $C_2H_6$ and $C_4H_8$). Therefore, $C_2H_2$ has the shortest C-C distance.

In a homologous series, each successive member differs by a -CH2- group. The molecular weight of a -CH2- group is 14 amu (12 for carbon and 2 for hydrogen). Therefore, the difference in atomic mass units (amu) between consecutive members of a homologous series is 14.

When potassium permanganate ($KMnO_4$) is passed through an acidified hydrogen sulfide ($H_2S$) solution, the $KMnO_4$ (a strong oxidizing agent) oxidizes the $H_2S$ to sulfur (S), while itself being reduced. The reaction is:

$2KMnO_4 + 3H_2SO_4 + 5H_2S → 2MnSO_4 + K_2SO_4 + 8H_2O + 5S$.

Therefore, the product formed is sulfur (S).

To determine the nature of the resulting solution, we must first calculate the milliequivalents of the acid and the base. For the acid: \[20 \, \text{ml} \times 0.25 \, N = 5 \, \text{milliequivalents}\]. For the base: \[30 \, \text{ml} \times 0.2 \, N = 6 \, \text{milliequivalents}\]. Since the base has more milliequivalents, the solution will be basic. To find the normality of the resulting solution: \[\text{Excess milliequivalents of base} = 6 - 5 = 1 \, \text{milliequivalent}\] and the total volume is \[20 + 30 = 50 \, \text{ml}\]. Thus, the normality is \[\frac{1}{50/1000} = 0.02 \, N\]. Since the base is in excess, the resulting solution is 0.02 N basic.

To determine the nature of the resulting solution, we calculate the number of millimoles of $H_2SO_4$ and NaOH. For $H_2SO_4$: \[10 \, \text{ml} \times 10 \, \text{M} \times 2 = 200 \text{millimoles of } H^+\] (since $H_2SO_4$ is diprotic). For NaOH: \[100 \, \text{ml} \times 1 \, \text{M} = 100 \text{millimoles of } OH^-\]. Since the millimoles of $H^+$ are greater than the millimoles of $OH^-$, the solution will be acidic.