A sample contains 1.25 grams of CH_4 and no other substances. What is the mole fraction of O_2 in this sample?

• A. 0.25
• B. 1
• C. 0
• D. 0.50

Answer: (C)

Verified step by step guidance

1. Identify the components present in the sample. The problem states the sample contains 1.25 grams of CH_4 and no other substances, so there is no O_2 present in the sample.
2. Recall that the mole fraction of a component in a mixture is given by the formula: \(\\chi_i = \frac{n_i}{\sum n_j}\), where \(n_i\) is the number of moles of component \(i\) and \(\sum n_j\) is the total number of moles of all components in the mixture.
3. Since the sample contains only CH_4 and no O_2, the number of moles of O_2 is zero: \(n_{O_2} = 0\).
4. Calculate the mole fraction of O_2 using the formula: \(\\chi_{O_2} = \frac{n_{O_2}}{n_{CH_4} + n_{O_2}} = \frac{0}{n_{CH_4} + 0} = 0\).
5. Therefore, the mole fraction of O_2 in the sample is zero because there is no oxygen present.