1. Use the ideal gas equation to find total moles

Given:

• Volume, $V = 2000\,\text{cm}^3 = 2 \times 10^{-3}\,\text{m}^3$
• Pressure, $P = 100\,\text{kPa} = 10^5\,\text{Pa}$
• Temperature, $T = 300\,\text{K}$
• Gas constant, $R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}$

Using $PV = nRT$ we get $n = \frac{PV}{RT} = \frac{10^5 \times 2\times 10^{-3}}{8.3 \times 300}$ $n = \frac{200}{2490} \approx 0.0803\,\text{mol}$

So, total number of moles in the mixture is $n_{\text{H}_2} + n_{\text{O}_2} = 0.0803$

2. Use the given mass of the mixture

Let:

• moles of hydrogen $= x$
• moles of oxygen $= y$

Then, $x + y = 0.0803$

Mass of mixture is $0.76\,\text{g}$.

Molar masses:

• $\text{H}_2 = 2\,\text{g/mol}$
• $\text{O}_2 = 32\,\text{g/mol}$

Hence, $2x + 32y = 0.76$

3. Solve the equations

From $x = 0.0803 - y$ Substitute into mass equation: $2(0.0803 - y) + 32y = 0.76$ $0.1606 - 2y + 32y = 0.76$ $30y = 0.5994$ $y \approx 0.0200$

Then, $x = 0.0803 - 0.0200 = 0.0603$

Thus, $\frac{x}{y} = \frac{0.0603}{0.0200} \approx 3$

So the ratio of number of moles of hydrogen to oxygen is $3:1$

4. Check options

• A: $\frac{1}{3}$ ❌
• B: $\frac{3}{1}$ ✅
• C: $\frac{1}{16}$ ❌
• D: $\frac{16}{1}$ ❌

Therefore, the correct answer is Option B.