no of moles of CHCl3 ( nCHCl3) = 0.838 moles no of moles of C2H6 ( n C2H6) = 0.1

Question

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

Explanation / Answer

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? 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?

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.