Asked by John
Chlorine is widely used to treat municipal water supplies and swimming pools. If the volume of a sample of Cl2 gas is 8.73L at 755mmHg and 24.0C, how many grams of Cl2 gas are in this sample?
This is one of those multi step problems that I get lost....I need to to know how to derive at an answer here....I think I have to convert units here or at least pay attention to the units somewhere...but I am getting lost with the calculation...Help!
This is one of those multi step problems that I get lost....I need to to know how to derive at an answer here....I think I have to convert units here or at least pay attention to the units somewhere...but I am getting lost with the calculation...Help!
Answers
Answered by
DrBob222
You must pay attention to units no matter what problem you are working.
Here, use PV = nRT and solve for n
P must be in atm and 755/760 = ? atm.
V in L
n = number of mols. Solve for this.
R = 0.08205 L*atm/mol*K
T = 24C + 273 = ?K. T must be in kelvin.
Then n = mols = grams/molar mass
You know molar mass Cl2 (about 71) and you know mols from PV = nRT, solve for grams.
Here, use PV = nRT and solve for n
P must be in atm and 755/760 = ? atm.
V in L
n = number of mols. Solve for this.
R = 0.08205 L*atm/mol*K
T = 24C + 273 = ?K. T must be in kelvin.
Then n = mols = grams/molar mass
You know molar mass Cl2 (about 71) and you know mols from PV = nRT, solve for grams.
Answered by
Tracy
n=PV/RT
n=(755/760)(8.73)÷24.37182
n=0.357mol
Mass=molar mass *number of moles
Mass=71g*0.357
=25.3g
n=(755/760)(8.73)÷24.37182
n=0.357mol
Mass=molar mass *number of moles
Mass=71g*0.357
=25.3g
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