At a certain temperature, the vapor pressure of pure thiophene ($C_4H_4S$) is measured to be 438 mmHg. Suppose a solution is prepared by mixing 98.3 g of thiophene and 79.0 g of heptane ($C_7H_{16}$). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits.

Note for advanced students: You may assume the solution is ideal.

The partial pressure of the thiophene vapor above the solution is calculated by applying Raoult's law. The mole fraction of thiophene in an ideal solution prepared with heptane is found, and then multiplied by the vapor pressure of pure thiophene to get the partial pressure.

Explanation:

To solve this problem, we first need to apply Raoult's law, which states that the partial pressure of a component in a solution is the product of the mole fraction of that component and the vapor pressure of the pure component.

First, we calculate the number of moles of each substance. The molar mass of thiophene (C4H4S) is approximately 84.14 g/mol and that of heptane (C7H16) is about 100.2 g/mol.

For Thiophene: Number of moles = 98.3 g / 84.14 g/mol = 1.17 mol
For Heptane: Number of moles = 79.0 g / 100.2 g/mol = 0.79 mol

Total moles in the solution = 1.17 mol + 0.79 mol = 1.96 mol. Now, we calculate the mole fraction of thiophene: mole fraction = moles of thiophene / total moles = 1.17 / 1.96 = 0.597.

Next, we apply Raoult's law to calculate the partial pressure of thiophene: Partial pressure of thiophene = mole fraction of thiophene x vapor pressure of pure thiophene = 0.597 x 438 mmHg = 260.7 mmHg (rounded to four significant digits).

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