To determine the amount of bromine needed to convert 2 g of phenol into 2,4,6-tribromophenol, we follow this procedure:
Step 1: Calculate the molar mass of phenol (C6H5OH)
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Molar mass of phenol = 6(12) + 6(1) + 16 = 94 g/mol
Step 2: Calculate the number of moles of phenol
Number of moles = mass / molar mass = 2 g / 94 g/mol ≈ 0.0213 mol
Step 3: Determine the reaction and molar mass of tribromophenol
Reaction: C6H5OH + 3 Br2 → C6H2Br3OH + 3 HBr
Molar mass of Br2 = 2(80) = 160 g/mol
Step 4: Calculate bromine required for the reaction
Bromine required per mol of phenol = 3 mols of Br2
Number of moles of Br2 = 3 × 0.0213 mol = 0.0639 mol
Mass of Br2 = moles × molar mass = 0.0639 mol × 160 g/mol = 10.224 g
Thus, the amount of bromine required is 10.22 g.
During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is %.
(Given molar mass in g mol\(^{-1}\) of Ba: 137, S: 32, O: 16)
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Two vessels A and B are connected via stopcock. Vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
Choose the correct nuclear process from the below options:
\( [ p : \text{proton}, n : \text{neutron}, e^- : \text{electron}, e^+ : \text{positron}, \nu : \text{neutrino}, \bar{\nu} : \text{antineutrino} ] \)
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: