QJEEM24Chaa0b: JEE Main 2024 Chemistry - Class 11 JEEM24Chaa0b Solution: Combustion of glucose ( C 6 H 12 O 6 ) produces CO 2 and water. …

To determine the amount of oxygen required for the complete combustion of 900 g of glucose, we need to follow these steps:

First, let's write the balanced chemical equation for the combustion of glucose $$({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6)$$:

$${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+6{\mathrm{O}}_2\to 6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$$

From the balanced equation, we see that 1 mole of glucose reacts with 6 moles of oxygen $$({\mathrm{O}}_2)$$.

Next, let's calculate the number of moles of glucose in 900 g. The molar mass of glucose is given as $$180\mathrm{g}{\mathrm{mol}}^{-1}$$.

Number of moles of glucose = $$\frac{900\mathrm{g}}{180\mathrm{g}{\mathrm{mol}}^{-1}}=5\mathrm{mol}$$

According to the balanced equation, 1 mole of glucose requires 6 moles of oxygen. Therefore, 5 moles of glucose will require:

Number of moles of oxygen = $$5\mathrm{mol}\times 6=30\mathrm{mol}$$

Now, we need to find the mass of 30 moles of oxygen. The molar mass of oxygen $$\left({\mathrm{O}}_2\right)$$ is $$32\mathrm{g}{\mathrm{mol}}^{-1}$$.

Mass of oxygen = $$30\mathrm{mol}\times 32\mathrm{g}{\mathrm{mol}}^{-1}=960\mathrm{g}$$

Therefore, the amount of oxygen required for the complete combustion of 900 g of glucose is 960 g.

So, the correct option is Option D: 960.