Home
Class 12
CHEMISTRY
At 300 K and 1 atm, 15 mL of a gaseous h...

At `300 K` and `1 atm, 15 mL` of a gaseous hydrocarbon requires `375 mL` air containing `20% O_(2)` by volume for complete combustion. After combustion, the gases occupy `330 mL`. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is

A

`C_(3)H_(8)`

B

`C_(4)H_(8)`

C

`C_(4)H_(10)`

D

`C_(3)H_(6)`

Text Solution

Verified by Experts

Promotional Banner

Topper's Solved these Questions

  • MOLE CONCEPT

    MOTION|Exercise EXERCISE - 1|64 Videos

  • MOLE CONCEPT

    MOTION|Exercise EXERCISE - 2 (LEVEL - I)|50 Videos

  • METALLURGY

    MOTION|Exercise EXERCISE 4 (LEVEL - II)|17 Videos

  • NOMENCLATURE OF ORGANIC COMPOUNDS

    MOTION|Exercise PREVIOUS YEAR|8 Videos

Similar Questions

Explore conceptually related problems

500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of CO_2 and 3 litres of water vapours. All volume being measured at the same temperature and pressure. The formula of the hydrocarbon is :

At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O_(2) for complete combustion, and 40 mL of CO_(2) is formed. The formula of the hydrocarbon is:

40 mL. of a hydrocarbon undergoes combustion in 260 ml of oxygen and gives 160 ml of carbondioxide. If all gases are measured under similar condtions of temperature and pressure, the formula of hydrocarbon is

2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion and produces 7.5 ml carbon dioxide and 10.0 ml water vapour. All the volumes are measured at the same pressure and temperature. Show that the data illustrates Gay Lussac’s law of volume combination.