Question

A gaseous mixture contains $8\text{ g}$ of oxygen, $14\text{ g}$ of nitrogen and $8\text{ g}$ of hydrogen. Total number of molecules present in the gaseous mixture is (Given: At. wt: $H=1$, $N=14$, $O=16$, $N_A = 6 \times 10^{23} \text{ mol}^{-1}$)

• A. $1.43 \times 10^{23}$
• B. $2.85 \times 10^{23}$
• C. $2.85 \times 10^{24}$
• D. $1.85 \times 10^{24}$

Hint:

Remember that common gases like Oxygen, Nitrogen, and Hydrogen are diatomic ($O_2, N_2, H_2$). A common mistake is using the atomic weights (16, 14, 1) instead of molecular weights (32, 28, 2) when calculating moles.

Answer 1

The Correct Option is C

Step 1: Calculate Moles of each Gas:
• Oxygen ($O_2$): Molar mass = $32\text{ g/mol}$. $$\text{Moles of } O_2 = \frac{8}{32} = 0.25\text{ mol}$$

• Nitrogen ($N_2$): Molar mass = $28\text{ g/mol}$. $$\text{Moles of } N_2 = \frac{14}{28} = 0.5\text{ mol}$$

• Hydrogen ($H_2$): Molar mass = $2\text{ g/mol}$. $$\text{Moles of } H_2 = \frac{8}{2} = 4.0\text{ mol}$$

Step 2: Calculate Total Moles: $$\text{Total Moles} = 0.25 + 0.5 + 4.0 = 4.75\text{ mol}\lt strong\gt Step 3: Calculate Total Molecules\lt /strong\gt \text{Total Molecules} = \text{Total Moles} \times N_A$$ $$\text{Total Molecules} = 4.75 \times 6 \times 10^{23}$$ $$\text{Total Molecules} = 28.5 \times 10^{23}$$ $$\text{Total Molecules} = 2.85 \times 10^{24}$$