Asked by Jude
The vapor pressure of pure water at 85 celsius is normally 433.6 mmHg but decrease to 393.8mmHg upon addition?
of an unknown amount of ammoniu chloride NH4Cl to 162.2g of water at this temperature. How many grams of ammonium chloride were added assuming complete dissociation of solute
of an unknown amount of ammoniu chloride NH4Cl to 162.2g of water at this temperature. How many grams of ammonium chloride were added assuming complete dissociation of solute
Answers
Answered by
DrBob222
delta P = XNH4Cl*P<sup>o</sup>
433.6-393.8 = XNH4Cl*433.6
XNH4Cl = 39.8/433.6 = ??
Then n = moles NH4Cl
162.2/18.015 = 9.00 moles H2O
2n/(2n+9) = 39.8/433.6 and solve for n.
Then grams NH4Cl = n NH4Cl * molar mass NH4Cl.
433.6-393.8 = XNH4Cl*433.6
XNH4Cl = 39.8/433.6 = ??
Then n = moles NH4Cl
162.2/18.015 = 9.00 moles H2O
2n/(2n+9) = 39.8/433.6 and solve for n.
Then grams NH4Cl = n NH4Cl * molar mass NH4Cl.
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