What is the partial pressure of neon in a 4.00 L vessel that contains 0.838 mol...

What is the partial pressure of neon in a 4.00 L vessel that contains 0.838 mol of chloroform, 0.184 mol of ethane, and 0.755 mol of neon at a total pressure of 928 torr?

Expert Solution

no of moles of CHCl3 ( nCHCl3) = 0.838 moles

no of moles of C2H6 ( n C2H6) = 0.184 moles

no of moles of Ne ( n Ne) = 0.755 moles

total no of moles = nCHCl3 + n C2H6 +n Ne

= 0.838+0.184+0.755 = 1.777 moles

mole fraction of neon ( X Ne) = n Ne/ nCHCl3 + n C2H6 +n Ne

= 0.755/1.777 = 0.425

partial pressure of neon = mole fraction of neon * total pressure

= 0.425*928 = 394.4torr >>>>>answer

queen_honey_blossom answered 2 years ago

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