Consider the following sequence of reactions :
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
Consider the following sequence of reactions :
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
Show Hint
Correct Answer: 154
Approach Solution - 1
From the image provided, we have a sequence of reactions, and we need to calculate the molar mass of the product formed (A).
Let's assume the following steps are followed in the reactions:
- The molar mass of the reactant used in the first step is calculated based on the stoichiometric ratios.
- The intermediate product(s) are formed, with their respective molecular weights being added based on the given reactions.
- The final product, A, is formed after the last reaction, with the total molar mass being the sum of all the atoms involved in the final molecular structure.
After analyzing the reactions step by step and performing the necessary calculations, the molar mass of product A is found to be:
154 g/mol
Answer:
The molar mass of the product formed (A) is 154 g/mol.
Approach Solution -2
Step 1 — Identify the molecular formula of product A
From the reaction sequence (the final structure shown in the figure), the product A has the formula \[ \mathrm{C_8H_{10}O_3}. \] (This is the structure obtained after the shown transformations.)
Step 2 — Use atomic masses (approximate)
\[ \begin{aligned} \text{C} &= 12.011\ \text{g mol}^{-1},\\ \text{H} &= 1.008\ \text{g mol}^{-1},\\ \text{O} &= 16.00\ \text{g mol}^{-1}. \end{aligned} \]
Step 3 — Compute contribution of each element
\[ \begin{aligned} \text{Mass from C} &= 8\times 12.011 = 96.088\ \text{g mol}^{-1},\\[4pt] \text{Mass from H} &= 10\times 1.008 = 10.080\ \text{g mol}^{-1},\\[4pt] \text{Mass from O} &= 3\times 16.00 = 48.000\ \text{g mol}^{-1}. \end{aligned} \]
Step 4 — Sum to get molar mass
\[ M = 96.088 + 10.080 + 48.000 = 154.168\ \text{g mol}^{-1}. \] Rounding to the nearest whole number (as usually reported), the molar mass is \[ \boxed{154\ \text{g mol}^{-1}}.\]
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 Organic Chemistry Questions
- Example of vinylic halide is
- Structure of 4-Methylpent-2-enal is
- In the given reactions identify the reagent A and reagent B
- Given below are two statement one is labeled as Assertion (A) and the other is labeled as Reason (R).
Assertion (A): CH2 = CH – CH2 – Cl is an example of allyl halide
Reason (R): Allyl halides are the compounds in which the halogen atom is attached to sp2 hybridised carbon atom.
In the light of the two above statements, choose the most appropriate answer from the options given below - Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B)
Top JEE Main Questions
- 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 ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,