Home
Class 11
CHEMISTRY
10 mL of H(2) combine with 5 mL of O(2) ...

10 mL of `H_(2)` combine with 5 mL of `O_(2)` to form `H_(2)O` when 200 mL of `H_(2)` at S.T.P. is passed through heated CuO, latter loses 0.144 g of its weight. Does the above data correspond to the law of constant composition ?

Text Solution

Verified by Experts

The correct Answer is:
Yes
In the first experiment :
Ratio by volume of `H_(2) and O_(2)` combining to form `H_(2)O=2 :1`
In the second experiment :
`CuO+H_(2)overset(heat)(rarr)Cu+H_(2)O`
The loss weight of CuO is due to oxygen which has been removed. It is 0.44 g
Now, 32 g of `O_(2)` at S.T.P occupy = 22400 L
0.144 g of `O_(2)` at S.T.P occupy `= (("22400 mL"))/(("32 g"))xx(0.144g)=100.8mL`
Ratio by volume of `H_(2) and O_(2)` combining to form `H_(2)O = 200 : 100.8 = 2:1`
Since the two ratios are the same, the data corresponds to Law of Constant Composition.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MOLE CONCEPT

    DINESH PUBLICATION|Exercise Competitive (MCQ)|45 Videos

  • MOLE CONCEPT

    DINESH PUBLICATION|Exercise JEE (Main)|31 Videos

  • MOLE CONCEPT

    DINESH PUBLICATION|Exercise Empirical molecular fromulae and chemical equations|8 Videos

  • HYDROGEN

    DINESH PUBLICATION|Exercise LONG|1 Videos

  • NOMENCLATURE OF ORGANIC COMPOUNDS

    DINESH PUBLICATION|Exercise MCQs|13 Videos

Similar Questions

Explore conceptually related problems

10 mL of hydrogen combine with 5mL of oxygen to yield water. When 200 mL of hydrogen at N.T.P. are passed over heated CuO , the latter loses 0.144 g of its mass. Do these results agree with the law of constant composition ?

10mL of hydrogen combine with 5 mL of oxygen to yield water. When 200 mL of hydrogen at NTP are passed over heated CuO, the CuO loses 0.144 g of its mass. Do these resutls correspond to the law of constant proportions?

Knowledge Check

  • 1 ml of H_(2)O_(2) solution given 10 ml of O_(2) at NTP. It is :

    A
    10 vol. `H_(2)O_(2)`
    B
    20 vol. `H_(2)O_(2)`
    C
    30 vol. `H_(2)O_(2)`
    D
    40 vol. `H_(2)O_(2)`

  • If 30mL of H_(2) and 20mL of O_(2) , reacts to form H_(2)O , what is left at the end of the reaction?

    A
    10mL of `H_(2)`
    B
    5mL of `H_(2)`
    C
    10mL of `O_(2)`
    D
    5mL of `O_(2)`

  • One ml. of H_(2)O_(2) solution gives 50 ml. of O_(2) at NTP, so it is

    A
    10 V
    B
    25 V
    C
    50 V
    D
    100 V

    • Similar Questions

    Explore conceptually related problems

    50 mL of ozone \( O_3 ) at STP were passed through 50 mL of 5 volume H_2O_2 solution. What is the volume strength of H_2O_2 after the reaction?

    One ml of H_(2)O_(2) solution gives 50ml of O_(2) at NTP, so it is

    If 30 mL of H_(2) and 20 mL of O_(2) reacts to form water, what is left at the end of the reaction?

    In 300 mL of a 5 volume H_(2)O_(2) sample, what mass of H_(2)O_(2) is there ?

    The volume of 10 volume of H_2O_2 required to liberate 500mL of O_2 at STP is

    Home
    Profile