A mixture of hydrogen and oxygen has volume 500 cm$$^3$$, temperature 300 K, pressure 400 kPa and mass 0.76 g. The ratio of masses of oxygen to hydrogen will be:

We are dealing with a gaseous mixture, so first of all we recall the ideal-gas equation

$$PV = nRT$$

where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the total number of moles present, $$R$$ is the universal gas constant and $$T$$ is absolute temperature.

All the data must be in SI units. We therefore convert each quantity one by one.

Pressure:
$$P = 400\ \text{kPa} = 400 \times 10^{3}\ \text{Pa} = 4 \times 10^{5}\ \text{Pa}$$

Volume:
$$V = 500\ \text{cm}^3 = 500 \times 10^{-6}\ \text{m}^3 = 5 \times 10^{-4}\ \text{m}^3$$

Temperature is already in kelvin, so
$$T = 300\ \text{K}$$

Universal gas constant:
$$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$

Substituting these values into $$PV = nRT$$ we get

$$n = \dfrac{PV}{RT} = \dfrac{\left(4 \times 10^{5}\ \text{Pa}\right)\left(5 \times 10^{-4}\ \text{m}^3\right)}{8.314\ \text{J mol}^{-1}\text{K}^{-1}\; \times\; 300\ \text{K}}$$

First multiply the numerator:

$$4 \times 10^{5}\; \text{Pa} \times 5 \times 10^{-4}\; \text{m}^3 = 200\ \text{J}$$

Next multiply the denominator:

$$8.314 \times 300 = 2494.2$$

Hence

$$n = \dfrac{200}{2494.2} \approx 0.08016\ \text{mol}$$

So the mixture contains in total

$$n_{\text{total}} = 0.08016\ \text{mol}$$

Let us denote the individual mole numbers as

$$n_{\text{O}_2} = x,\qquad n_{\text{H}_2} = y$$

Then we have the relation

$$x + y = 0.08016 \qquad\qquad (1)$$

The given total mass of the mixture is $$0.76\ \text{g}$$. Writing the mass in terms of molar masses, we obtain

$$32x + 2y = 0.76 \qquad\qquad (2)$$

To simplify equation (2) we divide every term by 2:

$$16x + y = 0.38 \qquad\qquad (3)$$

Now we solve the simultaneous equations (1) and (3).

From (1) we express $$y$$:

$$y = 0.08016 - x$$

Substituting this value of $$y$$ into (3):

$$16x + \left(0.08016 - x\right) = 0.38$$

Simplifying:

$$16x - x + 0.08016 = 0.38$$

$$15x + 0.08016 = 0.38$$

$$15x = 0.38 - 0.08016 = 0.29984$$

$$x = \dfrac{0.29984}{15} = 0.019989\ \text{mol} \approx 0.020\ \text{mol}$$

Using $$y = 0.08016 - x$$ gives

$$y = 0.08016 - 0.019989 = 0.060171\ \text{mol} \approx 0.060\ \text{mol}$$

Now we determine the individual masses.

Mass of oxygen:

$$m_{\text{O}_2} = 32x = 32 \times 0.019989 \approx 0.63965\ \text{g} \approx 0.64\ \text{g}$$

Mass of hydrogen:

$$m_{\text{H}_2} = 2y = 2 \times 0.060171 \approx 0.12034\ \text{g} \approx 0.12\ \text{g}$$

Hence the required ratio of the masses is

$$m_{\text{O}_2} : m_{\text{H}_2} = 0.64 : 0.12$$

Dividing both numbers by 0.04 gives

$$16 : 3$$

Hence, the correct answer is Option A.