Answer:
The correct answer is 81.52 percent.
Explanation:
Based on the given information, the boiling point of pentane is 36 degree C and the boiling point of ethanol is 78 degree C. The density of pentane and ethanol is 0.63 g/ml and 0.79 g/ml. The vapor pressure of pentane at 57.2 degree C is 1439 torr and the vapor pressure of ethanol at 57.2 degree C is 326 torr.
In the given case, 50 percent pentane by mass signifies that mass of pentane is 50 grams. Thus, the mass of ethanol will be 100-50 = 50 grams.
The moles or n can be calculated by using the formula,
n = weight/molecular mass
The molecular mass of pentane is 72.15 g per mol and the molar mass of ethanol is 46.07 g/mol.
The moles of pentane is,
= 50 g/72.15 g/mol = 0.6930 mol
The moles of ethanol is,
= 50 g/46.07 g/mol = 1.0853 mol
The mole fraction of pentane is,
= 0.6930 mol / (0.6930 + 1.0853) mol = 0.3897
The mole fraction of ethanol is,
= 1.0853 mol / (0.6930 + 1.0853) mol = 0.6103
Now the vapor pressure of solution will be,
= pressure of pentane * mole fraction of pentane + pressure of ethanol * mole fraction of ethanol
= (1439 * 0.3897) + (326 * 0.6103)
= 759.736 torr
The vapor pressure of pentane within the solution,
= vapor pressure of pentane * mole fraction of pentane
= 1439 torr * 0.3897
= 560.778 torr
The fraction of pentane is,
= 560.778 / 759.736 = 0.738
Let us assume that the total mole is 1, the mole fraction of pentane is 0.738, so the mole fraction of ethanol will become, 1-0.738 = 0.262
The mass of pentane = 0.738 * 72.15 = 53.2467
The mass of ethanol = 0.262 * 46.07 = 12.07034
The percent by mass of pentane in new solution will be,
Mass% = mass of pentane/Total mass * 100%
= 53.2467/(53.2467 + 12.07034) * 100%
= 53.2467/65.31704 * 100 %
= 81.52 %
