Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
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:
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
The molar mass of the product formed (A) is 154 g/mol.
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 \[ \boxed{154\ \text{g mol}^{-1}}.\]
The IUPAC name of the following compound is:
The compounds which give positive Fehling's test are:
Choose the CORRECT answer from the options given below:
The products formed in the following reaction sequence are: 
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to:
Given three identical bags each containing 10 balls, whose colours are as follows:
| Bag I | 3 Red | 2 Blue | 5 Green |
| Bag II | 4 Red | 3 Blue | 3 Green |
| Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is:
If \( \theta \in \left[ -\frac{7\pi}{6}, \frac{4\pi}{3} \right] \), then the number of solutions of \[ \sqrt{3} \csc^2 \theta - 2(\sqrt{3} - 1)\csc \theta - 4 = 0 \] is equal to ______.