A mixture of hydrogen and oxygen has volume $$2000$$ cm$$^3$$, temperature $$300$$ K, pressure $$100$$ kPa and mass $$0.76$$ g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be
[Take gas constant $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$]

Given: $$V = 2000 \text{ cm}^3 = 2 \times 10^{-3} \text{ m}^3$$, $$T = 300$$ K, $$P = 100$$ kPa $$= 10^5$$ Pa, total mass $$= 0.76$$ g, $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$.

Let the number of moles of hydrogen be $$n_1$$ and oxygen be $$n_2$$.

Using the ideal gas law for the mixture:

$$PV = (n_1 + n_2)RT$$

$$10^5 \times 2 \times 10^{-3} = (n_1 + n_2) \times 8.3 \times 300$$

$$200 = (n_1 + n_2) \times 2490$$

$$n_1 + n_2 = \frac{200}{2490} = \frac{20}{249} \approx 0.08 \text{ mol}$$

From the total mass:

$$2n_1 + 32n_2 = 0.76$$ g

From the ideal gas equation: $$n_1 + n_2 = 0.08$$, so $$n_2 = 0.08 - n_1$$.

Substituting:

$$2n_1 + 32(0.08 - n_1) = 0.76$$

$$2n_1 + 2.56 - 32n_1 = 0.76$$

$$-30n_1 = 0.76 - 2.56 = -1.80$$

$$n_1 = \frac{1.80}{30} = 0.06 \text{ mol}$$

$$n_2 = 0.08 - 0.06 = 0.02 \text{ mol}$$

Therefore:

$$\frac{n_1}{n_2} = \frac{0.06}{0.02} = \frac{3}{1}$$

The correct answer is Option B.