The mole ratio of H₂ and O₂ gas is 8:1. The ratio of their weight will be:

1. 1 : 1
2. 2 : 1
3. 4 : 1
4. 1 : 2

The specific volume of cylindrical virus particle is $6.02$ $\times$ ${10}^{-2}$ $cc/g\mathrm{m\; w}$hose radius and length are 7 Å & 10Å respectively. If $N_A$ $=$ $6.023$ $\times {10}^{23}$\(\text{mol}^{-1}\), find the molecular weight.

1. \( 15.4 \mathrm{~kg} / \mathrm{mol} \)
2. \(1.54 \times 10^4 \mathrm{~kg} / \mathrm{mol} \)
3. \( 3.08 \times 10^4 \mathrm{~kg} / \mathrm{mol}\)
4. \(3.08 \times 10^3 \mathrm{~kg} / \mathrm{mol}\)

Which among the following has the maximum number of molecules?

1. 7 gm N₂

2. 2 gm H₂

3. 16 gm NO₂

4. 16 gm O₂

The maximum number of molecules is present in: