Home
Class 12
CHEMISTRY
The ratio of number of oxygen atoms (O) ...

The ratio of number of oxygen atoms (O) in 16.0 g ozone `(O_(3))`, 28.0 g carbon monoxide (CO) and 16.0 g oxygen `(O_(2))` is :-
(Atomic mass : C = 12, O = 16 and Avogadro's constant `N_(A) = 6.0 xx 10^(23) mol^(-1))`

A

`3 : 1 : 1`

B

`1 : 1 : 2`

C

`3 : 1 : 2`

D

`1 : 1 : 1`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • MOLE CONCEPT

    ALLEN|Exercise J-ADVANCE|2 Videos

  • MOLE CONCEPT

    ALLEN|Exercise O-II (Match the column:)|3 Videos

  • METALLURGY

    ALLEN|Exercise EXERCISE-05 [B]|26 Videos

  • p-Block Element

    ALLEN|Exercise All Questions|20 Videos

Similar Questions

Explore conceptually related problems

The ratio of number of oxygen atoms (O) in 16.0g ozone (O_(3)) . 28.0g carbon monoxide (CO) and 16.0g oxygen (O_(2)) is :- ("Atomic mass" : C=12,O=16 "and Avogadro's constant" N_(A) =6.0xx10^(23) "mol"^(-1))

Convert 22 g of carbon dioxide (CO_(2)) into moles. (Atomic masses : C = 12 u, O = 16 u )

Which contains more molecules, 10 g of sulphur dioxide (SO_(2)) or 10 g of oxygen (O_(2)) ? (Atomic masses : S = 32 u , O = 16 u )

Which contains more molecules, 4 g of methane (CH_(4)) or 4g of oxyen (O_(2)) ? (Atomic masses : C = 12 u, H = 1 , O = 16 u )

8 g of O_(2) has the same number of oxygen atoms as