Respuesta :
Answer:
The answers are:
Partial pressure Xenon = 174.3 mmHg
Partial pressure Nitrogen = 608.7 mmHg
Explanation:
According to Dalton's law, the partial pressure of a gas in a mixture is equal to the total pressure multiplied by the molar fraction of the gas:
P= X Ptotal
In this case we hace a mixture of xenon (Xe) and nitrogen (N₂) gases. In order to calculate the partial pressures of each gas, we have to first calculate the molar fraction of each one:
X= number of moles of gas/total moles
Xenon is a monoatomic gas (molecular weight= molar mass= 131.3 g/mol) and nitrogen is a diatomic gas (molecular weight= molar mass N x 2= 14 g/mol x 2= 28 g/mol)
moles Xe= mass Xe/molecular weight Xe= 9 g/131.3 g/mol= 0.068 mol
moles N₂= mass N₂/molecular weight N₂= 6.65 g/28 g= 0.2375 mol
Total moles= moles Xe + moles N₂ = 0.068 mol + 0.2375 mol = 0.3055 mol
⇒X(Xe)= moles Xe/total moles= 0.068 mol/0.3055= 0.2226
P(Xe)= X(Xe) x Ptotal = 0.2225 x 783 mmHg = 174.3 mmHg
⇒X(N₂)= moles N₂/total moles= 0.2375 mol/0.3055= 0.7774
P(Xe)= X(Xe) x Ptotal = 0.7774 x 783 mmHg = 608.7 mmHg
Answer:
Partial pressure Xe = 175.4 mm Hg
Partial pressure N2 = 607.6 mm Hg
Explanation:
Step 1: Data given
Total pressure of the mixture = 783 mm Hg
Mass of Xenon = 9.00 grams
Atomic mass Xenon = 131.29 g/mol
Mass of N2 gas = 6.65 grams
Molar mass of 28.0 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles xenon = 9.00 grams / 131.29 g/mol
Moles Xenon = 0.06855 moles
Moles nitrogen gas = 6.65 grams / 28.0 g/mol
Moles nitrogen gas = 0.2375 moles
Step 3: Calculate mol ratio
Mol ratio = number of moles / total moles
Mol ratio xenon = 0.06855 / (0.06855+0.2375)
Mol ratio xenon = 0.224
Mol ratio nitrogen gas = 0.2375 / (0.06855+0.2375)
Mol ratio nitrogen gas = 0.776
Step 4: Calculate the partial pressures
Partial pressure = mol ratio * total pressure
Partial pressure Xe = 0.224 * 783 mmHg
Partial pressure Xe = 175.4 mm Hg
Partial pressure N2 = 0.776 * 783 mmHg
Partial pressure N2 = 607.6 mm Hg
175.4 + 607.6 = 783 mmHg
