NEET Practice Questions (MCQs) with Answers & Solutions

To find the percentage of C and H in the compound, we need to perform the following calculations:

1. Calculate the mass of carbon in CO$_2$: \[ \text{Mass of C} = \frac{12}{44} \times 0.198 \text{ g} = 0.054 \text{ g} \]

2. Calculate the mass of hydrogen in H$_2$O: \[ \text{Mass of H} = \frac{2}{18} \times 0.1014 \text{ g} = 0.0113 \text{ g} \]

3. Calculate the percentage of C and H in the organic compound: \[ \%C = \frac{0.054}{0.246} \times 100 = 21.95\% \] \[ \%H = \frac{0.0113}{0.246} \times 100 = 4.58\% \]

Hence, the correct percentages of C and H are 21.95% and 4.58%, respectively. Therefore, the correct option is 21.95, 4.58.

Both the Dumas and Kjeldahl methods are used for estimating nitrogen in organic compounds. The Dumas method involves combustion of the compound and measuring the nitrogen gas released, while the Kjeldahl method involves digestion of the compound with a strong acid and measuring the ammonia produced.

In the Dumas method, the gaseous mixture produced is passed through aqueous potassium hydroxide (KOH). The $CO_2$ in the gaseous mixture is absorbed by the KOH, allowing nitrogen gas to be measured accurately.

In Dumas method, the formula for calculating the percentage of nitrogen (% N) in an organic compound is given by: \[ \% N = \frac{28 \times V \times 100}{22400 \times m} \] where:

• $V$ is the volume of nitrogen gas in liters,
• $m$ is the mass of the organic compound in grams. This formula allows for the determination of the nitrogen content based on the volume of nitrogen gas produced.

P = 715 - 15 = 700 mm $ Vol of N_2 at STP = { 273 \times 700 \times 50 \over 300 \times 760} = 41.9 ml $ $ \therefore \% N = { 28 \times 41.9 \times 100 \over 22400 \times 0.3 } = 17.46 \% $

In Kjeldahl’s method, the nitrogenous organic compound is heated with concentrated sulfuric acid. During this process, nitrogen is converted into ammonium sulfate, $(NH_4)_2SO_4$. The reaction can be represented as: \[ \text{Organic compound} + H_2SO_4 \rightarrow (NH_4)_2SO_4 \] This conversion is crucial for the subsequent determination of nitrogen content.

$ 1M of 10 ml H_2SO_4 = 1 M of 20ml NH_3 $ $ 1000ml of 1M NH_3 = 14 g nitrogen $ $ 20 ml of 1M NH_3 ={14 \times 20 \over 1000} g nitrogen $ $ %N = { 14 \times 20 \times 100 \over 1000 \times 0.5 } = 56.0% $

In the Carius method for halogen estimation, the halogen present in the organic compound is converted into silver halide (AgX) by reacting with silver nitrate (AgNO_3). The reaction is as follows: \[ \text{R-X} + AgNO_3 \rightarrow AgX + \text{R-NO_3} \] where R-X is the organic compound containing the halogen. Silver halide precipitates out and can be weighed to determine the halogen content.