Suppose you dissolved 0.123 gram of pentane in 2.493 grams of p-xylene and measured a freezing point depression of 2.88 degrees celcius for the solution. Calculate the molar mass of pentane using this data and the value for Kf that you calculated in question 1 I got .829 mol/kg for question 1
Roult’s Law states that the vapor pressure, p, of a solution of a non-volatile solute is equal to the vapor pressure of the pure solvent, Po solv, times the mole fraction of the solvent, Xsolv
p = Xsolv * Po sol
X solv = number of moles of solvent / number of moles of solution
The solvent is water and the solute (not volatile) is glycerin.
Number of moles = mass in grams / molar mass
mass of water = 132 ml * 1 g/ml = 132 g
molar mass of water = 18 g/mol
=> number of moles of water = 132 g / 18 g/mol = 7.33333 mol
mass of glycerin = 27.2 g
molar mass of glycerin:, C3H8O3: 3 * 12 g/mol + 8 * 1 g/mol + 3*16 g/mol = 92 g/mol
number of moles of glycerin = 27.2g / 92 g/mol = 0.29565
total number of moles = 7.33333 moles + 0.29565 moles = 7.62898 moles
=> X solv = 7.33333 / 7.62898 = 0.96125
=> p = 0.96125 * 31.8 torr ≈ 30.57 torr ≈ 30.6 torr.
Answer: 30.6 torr
|Category||Business & Management|
|Paper Type||Case Study Writing|
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