Question
In: Chemistry
A solution consists of 3.88 g benzene,C6H6 , and 2.45 g toluene ,C6H5CH3. The vapor pressure...
A solution consists of 3.88 g benzene,C6H6 , and 2.45 g toluene ,C6H5CH3. The vapor pressure of pure benzene at 20. °C is 75 mm Hg and that of toluene at 20.0 °C is 22 mm Hg. Assume that Raoult’s lawholds for each component of the solution, calculate the mole fraction of benzene in the vapor. ( molar mass of benzene= 78.0 g/mole and toluene = 92.0 g/mole.)
Answer= 0.87
Solutions
Expert Solution
moles of benzene = nB = 3.88 / 78 = 0.0497
moles of toulene = nT = 2.45 / 92 = 0.0266
total moles = nB + nT = 0.0763
moles fraction of benzene in liquid phase XB = moles of benzene / total moles = 0.0497/ 0.0763 = 0.65
moles fraction of toulene in liquid phase XT = 0.35
total pressure = XB Po + XT Po
= 0.65 x 75 + 0.35 x 22
= 56.45 mmHg
P = total = 56.45 mmHg
mole fraction of benzene in vaphor phase = YB
XB PB = YB P
0.65 x 75 = YB x 56.45
YB = 0.87
mole fraction of benzene in vaphor phase = YB = 0.87
queen_honey_blossom answered 2 years agoRelated Solutions
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Tmelting = 279.0 K Tboiling = 353.0 K
1) Determine number of moles the sample contains.
___________________moles
2) How much energy must be removed to...
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