i have been working on this problem for a long time to no avail. ANY direction would be appreciated because i keep hitting a wall. thank you!!
The vapor pressure of benzene is 73.03 mm Hg at 25oC.
How many grams of TNT (trinitrotoluene), C7H5N3O6, a nonvolatile, nonelectrolyte (MW = 227.1 g/mol), must be added to 172.0 grams of benzene to reduce the vapor pressure to 72.04 mm Hg ?
benzene = C6H6 = 78.12 g/mol.
The vapor pressure of benzene is 73.03 mm Hg at 25oC.
How many grams of TNT (trinitrotoluene), C7H5N3O6, a nonvolatile, nonelectrolyte (MW = 227.1 g/mol), must be added to 172.0 grams of benzene to reduce the vapor pressure to 72.04 mm Hg ?
benzene = C6H6 = 78.12 g/mol.
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72.04 / 73.03 = 0.9864 = mole fraction of benzene.
172.0g x [1mole / 78.12g] = 2.202moles benzene.
2.202 / 0.9864 = 2.232 moles total.
2.232 - 2.202 = 0.030 moles TNT
0.030 moles x [227.2g / 1mole ] = 6.8g TNT. (to 2 digits like 0.030)
172.0g x [1mole / 78.12g] = 2.202moles benzene.
2.202 / 0.9864 = 2.232 moles total.
2.232 - 2.202 = 0.030 moles TNT
0.030 moles x [227.2g / 1mole ] = 6.8g TNT. (to 2 digits like 0.030)