What amount of bromine will be required to convert 2 g of phenol into 2, 4, 6-tribromophenol?
(Given molar mass in g mol\(^{-1}\) of C, H, O, Br are 12, 1, 16, 80 respectively)
Show Hint
- 10.22 g
- 6.0 g
- 4.0 g
- 20.44 g
The Correct Option is A
Approach Solution - 1
To determine the amount of bromine required to convert 2 g of phenol into 2,4,6-tribromophenol, we need to consider the balanced chemical reaction and the stoichiometry involved. Let's go through the steps:
- The reaction between phenol \( \text{C}_6\text{H}_5\text{OH} \) and bromine (\( \text{Br}_2 \)) to form 2,4,6-tribromophenol \( \text{C}_6\text{H}_2\text{Br}_3\text{OH} \) is as follows:
\(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\) - Determine the molar masses:
- Molar mass of phenol (\( \text{C}_6\text{H}_5\text{OH} \)):
Carbon (C): 12 × 6 = 72 g/mol
Hydrogen (H): 1 × 6 = 6 g/mol
Oxygen (O): 16 × 1 = 16 g/mol
Total = 72 + 6 + 16 = 94 g/mol - Molar mass of bromine (\( \text{Br}_2 \)):
Bromine (Br): 80 × 2 = 160 g/mol
- Molar mass of phenol (\( \text{C}_6\text{H}_5\text{OH} \)):
- Determine the moles of phenol:
\(n_{\text{phenol}} = \frac{2 \text{ g}}{94 \text{ g/mol}} \approx 0.0213 \text{ mol}\) - According to the balanced equation, 1 mole of phenol reacts with 3 moles of bromine. Therefore, the moles of bromine required will be:
\(n_{\text{Br}_2} = 3 \times 0.0213 = 0.0639 \text{ mol}\) - Calculate the mass of bromine needed:
\(m_{\text{Br}_2} = 0.0639 \text{ mol} \times 160 \text{ g/mol} = 10.224 \text{ g}\)
Therefore, the amount of bromine required is approximately 10.22 g, corresponding to the correct option: 10.22 g. This calculation confirms that option \(10.22 \text{ g}\) is correct.
Approach Solution -2
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.
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Stoichiometry and Stoichiometric Calculations Questions
- If 50 mL of 0.5 M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _____g.
- Consider the following reaction:\[3\text{PbCl}_2 + 2(\text{NH}_4)_3\text{PO}_4 \rightarrow \text{Pb}_3(\text{PO}_4)_2 + 6\text{NH}_4\text{Cl}\]If 72 mmol of PbCl\(_2\) is mixed with 50 mmol of (\(\text{NH}_4\))\(_3\)PO\(_4\), then the amount of \(\text{Pb}_3(\text{PO}_4)_2\) formed is ______ mmol (nearest integer).
- \( X \, \text{g} \) of ethylamine is subjected to reaction with \( \text{NaNO}_2/\text{HCl} \) followed by water; evolved dinitrogen gas which occupied \( 2.24 \, \text{L} \) volume at STP. \( X \) is ______ \( \times 10^{-1} \, \text{g} \).
- 9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product 'P'. The mass of product 'P' obtained is 26.4 g. The percentage yield is ......... %.
- Combustion of glucose (\(C_6H_{12}O_6\)) produces \(CO_2\) and water. The amount of oxygen (in g) required for the complete combustion of \(900 \, \text{g}\) of glucose is:
\([ \text{Molar mass of glucose in g mol}^{-1} = 180 ]\)
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.