JEE Main | 2026SolutionMedium
Question
80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
Options
A.[image/equation - edit required]
B.[image/equation - edit required]
C.[image/equation - edit required]
D.[image/equation - edit required]
Solution
Detailed solution available for registered users.
Create a free account to view solution
View Solution FreeMore Solution Questions
An aqueous solution of non-volatile and nonelectrolyte solute boils at 100.78°C. It should freeze at (for water, Kf and ...The vapour pressure of a solvent is decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mo...In which case, van′t Hoff factor i remains unchanged ? [Assume common complexes of these ions]...When 36.0 g of a solute having the empirical formula CH2O is dissolved in 1.20 kg of water, the solution freezes at −0.9...Liquids A and B form an ideal solution. A certain solution of A and B contains 25 mole percent of A, whereas the vapours...