1. Use the ideal gas equation to find total moles

Given:

• Volume: $V = 500\,\text{cm}^3 = 5 \times 10^{-4}\,\text{m}^3$
• Temperature: $T = 300\,\text{K}$
• Pressure: $P = 400\,\text{kPa} = 4 \times 10^5\,\text{Pa}$
• Total mass of mixture: $m = 0.76\,\text{g}$

Using $PV = nRT$

$n = \frac{PV}{RT} = \frac{(4\times 10^5)(5\times 10^{-4})}{8.314 \times 300}$

$n = \frac{200}{2494.2} \approx 0.0802\,\text{mol}$

So total moles of the gas mixture are approximately $n \approx 0.08\,\text{mol}$

2. Find average molar mass of the mixture

$M_{\text{avg}} = \frac{\text{total mass}}{\text{total moles}} = \frac{0.76}{0.08} = 9.5\,\text{g mol}^{-1}$

3. Let masses of hydrogen and oxygen be $m_H$ and $m_O$

Then $m_H + m_O = 0.76$

Moles of hydrogen gas $H_2$: $n_H = \frac{m_H}{2}$

Moles of oxygen gas $O_2$: $n_O = \frac{m_O}{32}$

Total moles: $\frac{m_H}{2} + \frac{m_O}{32} = 0.08$

Also, using $m_H = 0.76 - m_O$:

$\frac{0.76 - m_O}{2} + \frac{m_O}{32} = 0.08$

Multiply by 32:

$16(0.76 - m_O) + m_O = 2.56$

$12.16 - 16m_O + m_O = 2.56$

$12.16 - 15m_O = 2.56$

$15m_O = 9.60$

$m_O = 0.64\,\text{g}$

Then $m_H = 0.76 - 0.64 = 0.12\,\text{g}$

4. Find the ratio of masses

$m_O : m_H = 0.64 : 0.12 = 64 : 12 = 16 : 3$

5. Check options

• A: $3:8$ ❌
• B: $3:16$ ❌
• C: $16:3$ ✅
• D: $8:3$ ❌

Therefore, the correct option is C.