NEET Practice Questions (MCQs) with Answers & Solutions

In the Lassaigne test for the detection of Sulphur, a black precipitate is formed due to the formation of lead sulfide (PbS). When the sodium fusion extract (Lassaigne's extract) containing sulfur is treated with lead acetate, a black precipitate of lead sulfide is obtained. The reaction can be represented as: \[ Na_2S + (CH_3COO)_2Pb \rightarrow PbS + 2CH_3COONa \\]

When a Lassaigne solution containing sulfur is treated with sodium nitroprusside, a violet color is observed. This is because sulfur present in the solution reacts with sodium nitroprusside to form a violet-colored complex. The reaction can be represented as: \[ Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS] \\]

The violet color observed in the Lassaigne test for sulfur is due to the formation of the complex ion \[ [Fe(CN)_5NOS]^{4-} \\]. This complex is formed when the sulfur in the sample reacts with sodium nitroprusside in the solution. The reaction can be represented as: \[ Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS] \\]

During the estimation of Carbon (C) and Hydrogen (H), the carbon dioxide ($CO_2$) produced is absorbed in potassium hydroxide (KOH). This is because KOH reacts with $CO_2$ to form potassium carbonate ($K_2CO_3$), effectively removing $CO_2$ from the mixture.

During the estimation of Carbon (C) and Hydrogen (H), the water ($H_2O$) produced is absorbed in anhydrous calcium chloride ($An.CaCl_2$). Anhydrous calcium chloride is a highly hygroscopic substance, meaning it readily absorbs moisture from its surroundings.

In Kjeldahl’s method for the estimation of nitrogen, the formula used is $ \\% N = { 1.4 NV \over W} $. Here, N is normality of acid, V is volume of acid, and W is the weight of the sample. This formula helps in calculating the percentage of nitrogen in the given sample.

Elements No. of moles Simple ratio C 6/12 = 0.5 1 H 1/1 = 1 2 O 8/16 = 0.5 1 $ \therefore E.F = CH_2O $ M.W = 2VD =60 $ \therefore n = 2 \therefore M.F = C_2H_4O_2 $

$ \% S = { 32 \over 233} \times { 0.35 \over 0.2595} \times 100 = 18.52 $

The incorrect formula among the given options is: \( \%C = \frac{12 \times m \times 100}{44 \times V} \)

Explanation: The correct formula to calculate the percentage of carbon is \(\%C = \frac{12 \times m}{44 \times V} \times 100\), where \(m\) is the mass of CO$_2$ and \(V\) is the volume of the gas. The given formula includes an extra multiplication by \(m\) and hence is incorrect.