NEET + JEEAdvanced
A sealed vessel contains a mixture of carbon monoxide (CO, molar mass…
Question
A sealed vessel contains a mixture of carbon monoxide (CO, molar mass 28) and carbon dioxide (CO2, molar mass 44) gases. The total amount of gas is 0.500 mol and the total mass of the mixture is 18.0 g. Calculate the mole fraction of CO2 in the mixture.
✓ Verified answer: 0.5checked by our engine — not a guess
Step-by-step solution
Let y = mol CO, so mol CO2 = 0.500 - y.
28y + 44(0.500 - y) = 18.0 -> 28y + 22 - 44y = 18 -> -16y = -4 -> y = 0.250 mol CO.
Mol CO2 = 0.500 - 0.250 = 0.250.
Mole fraction CO2 = 0.250/0.500 = 0.500.
Final answer0.5
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Phys Chem Mole solutions
A 5.00 g mixture of calcium carbonate (CaCO3, molar mass 100 g/mol)…NEET + JEE · PhysicsA hydrated salt MgSO4.xH2O (anhydrous MgSO4 molar mass 120 g/mol,…NEET + JEE · PhysicsA 4.60 g sample of a pure organic compound containing only carbon,…NEET + JEE · PhysicsIn the Haber process, N2 + 3H2 -> 2NH3. A reaction vessel is charged…NEET + JEE · PhysicsConcentrated sulphuric acid is 98.0% H2SO4 by mass (molar mass 98…NEET + JEE · PhysicsBoron occurs as two isotopes of masses exactly 10 u and 11 u. The…NEET + JEE · Physics6.5 g of zinc (molar mass 65 g/mol) is added to excess dilute…NEET + JEE · PhysicsA 25.0 mL sample of phosphoric acid (H3PO4) solution is exactly…NEET + JEE · Physics