Home
Class 11
CHEMISTRY
A definite amout of gaseous hydrocarbon ...

A definite amout of gaseous hydrocarbon was burnt with just sufficient of `O_(2)`. The volume of all reactants was `600ml`, after the explosion the volume of the products `[CO_(2)(g)` and `H_(2)O(g)` was found to be `700ml` under the similar conditions the molecular formula of the compound is .

A

`C_(3)H_(8)`

B

`C_(3)H_(6)`

C

`C_(3)H_(4)`

D

`C_(4)H_(10)`

Text Solution

Verified by Experts

The correct Answer is:
A
Mass `%` of `O=((60/(98)xx4xx16+(40)/(80)xx3xx16)xx100)/(100) = 63.18%`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MOLE CONCEPT

    ALLEN|Exercise EXERCISE - 02|46 Videos

  • MOLE CONCEPT

    ALLEN|Exercise EXERCISE -03|14 Videos

  • MOLE CONCEPT

    ALLEN|Exercise Exercise - 05(B)|8 Videos

  • IUPAC NOMENCLATURE

    ALLEN|Exercise Exercise - 05(B)|8 Videos

  • QUANTUM NUMBER & PERIODIC TABLE

    ALLEN|Exercise J-ADVANCED EXERCISE|26 Videos

Similar Questions

Explore conceptually related problems

8.4 ml of a gaseous hydrocarbon (A) was burnt with 50 ml of O_(2) in an eudiometer tube. The volume of the products after cooling to room temperature was 37.4 ml, when reacted with NaOH, the volume contracted to 3.8 ml. What is the molecular formula of A?

A sample of gaseous hydrocarbon occupying 1.12 litres at NTP when completely burnt in air produced 2.2g of CO_(2) and 1.8g of H_(2)O . Calculate the weight of the compound and the volume of the oxygen at NTP required for its burning. Find the molecular formula of the compound

Knowledge Check

  • A definite amount of gaseous hydrocarbon having ("carbon atoms less than" 5) was burnt with sufficient amount of O_(2) . The volume of all reactants was 600mL . After the explosion the volume of the product [CO_(2)(g) and H_(2)O (g)] was found to be 700 mL under the similar conditions. The molecular formula of the compound is ?

    A
    `C_(3)H_(8)`
    B
    `C_(3)H_(6)`
    C
    `C_(3)H_(4)`
    D
    `C_(4)H_(10)`

  • 500 ml of a gaseous hydrocarbon when burnt in excess of O_(2) gave 2.0litres of CO_(2) and 2.5 litres of water vapours under same conditions. molecular formula of the hydrocarbon is:-

    A
    `C_(4)H_(8)`
    B
    `C_(4)H_(10)`
    C
    `C_(5)H_(10)`
    D
    `C_(5)H_(12)`

  • 10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " O_(2) and exploded under condition which allowed the H_(2)O formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition. Q.The molecular formula of the compound is

    A
    `C_(3)H_(8)O`
    B
    `C_(3)H_(6)O`
    C
    `C_(2)H_(6)O`
    D
    `C_(2)H_(4)O`

    • Similar Questions

    Explore conceptually related problems

    8.4 " mL of " gaseous hydrocarbon A was burnt with 50 " mL of " O_2 in a eudiometer tube. The volume of the products after cooling to room temperature was 37.4 mL. When reacted with NaOH, the volume contracted to 3.8 mL. What is the molecular formula of A.

    10 ml of gaseous hydrocarbon on combution gives 20ml of CO_(2) and 30ml of H_(2)O(g) . The hydrocarbon is :-

    8.4 ml of a gaseous hydrocarbon was burnt with 50 ml of O_2 in a eudiometer tube. The volume of the products formed after cooling to room temperature was 37.4 ml, when reacted with NaOH, the volume contracted to 3.8 ml. The molecular formula of the hydrocarbon is

    One volume of a gaseous organic compound of C, H and N on combustion produced 2 volumes of CO_(2) , 3.5 volumes of H_(2)O vapour and 0.5 volume of N_(2) under identical conditions of temperature and pressure. The molecular formula of the compound is

    Ten millilitre of a gaseous hydrocarbon was burnt completely in 80 ml of O_2 at STP. The volume of the remaining gas is 70 ml. The volume became 50 ml, on treatment with NaOH . The formula of the hydrocarbon is:

    Home
    Profile