Let us analyze each statement:
Incorrect. Glucose is soluble in water due to the presence of multiple hydroxyl ($-$OH) groups, not because of the aldehyde functional group. The hydroxyl groups form hydrogen bonds with water, making glucose highly soluble.
Correct. In aqueous solution, glucose exists in equilibrium between its open-chain form and cyclic isomeric forms ($\alpha$- and $\beta$-D-glucose).
Correct. Glucose is classified as an aldohexose because it contains six carbon atoms (hexose) and an aldehyde group (aldo).
Correct. Glucose is one of the two monomeric units of sucrose, the other being fructose.
Final Answer: (1)
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]