Home
Class 12
CHEMISTRY
The abundance of three isotopes of oxyge...

The abundance of three isotopes of oxygen (atomic mass 16.12) contains 8, 9, 10 neutrons respectively. One of the heaviest isotopes has 2% abundance. The other two are :

A

90, 8

B

80, 18

C

60, 38

D

18, 80

Text Solution

Verified by Experts

The correct Answer is:
A
`16.12 = (16 xx a + 17 xx (98-a) + 18 xx 2)/(100) " "therefore a = 90% "for"_(16)O and 8% "for"^(17)O`.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Multiple Correct Answer Type) |10 Videos

  • NUCLEAR CHEMISTRY

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Numerical type) |6 Videos

  • NUCLEAR CHEMISTRY

    BRILLIANT PUBLICATION|Exercise LEVEL-II|38 Videos

  • NITROGEN COMPOUNDS

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Linked Comprehension Type)|9 Videos

  • P-BLOCK ELEMENTS

    BRILLIANT PUBLICATION|Exercise LEVEL-III|50 Videos

Similar Questions

Explore conceptually related problems

If the average atomic mass of an element having two isotopes is 71, the ratio of the natural abundance of the two isotopes of mass 70 and 74 will be

A and B are two isotopes having mass numbers 14 and 16 respectively. If the atomic number of A is 7, then how many neutrons will be there in B?

Knowledge Check

  • 1 mole of a diatomic element X_2 contains 34 and 40 moles of electrons and neutrons respectively. The isotopic formula of the element is:

    A
    `""_(34)^(74)X`
    B
    `""_(17)^(37)X`
    C
    `""_(34)^(40)X`
    D
    `""_(20)^(40)X`

  • In each of the following question, an Assertion (A) is following by a corresponding Reason (R ). Use the following keys to choose the appropriate answer. Assertion : Average atomic mass of an element depends upon the atomic masses of different isotopes and their fractional abundances. Reason : Average atomic mass is obtained by multiplying the atomic mass of the heavier isotope with its fractional abundances.

    A
    If both (A) and ( R ) are correct and ( R ) is the correct explanation of (A).
    B
    If both (A) and (R ) are correct, but ( R ) is not the correct explanation of (A).
    C
    If (A) is correct, but ( R) is incorrect.
    D
    If both (A) and (R ) are incorrect.

  • Statement 1 : Average atomic mass of chlorine is 35.5 amu. Statement 2 : Chlorine has two isotopes Cl-35 and Cl-37 and their relative abundance is 3:1.

    A
    Statement 1 is True, statement 2 is True, Statement 2 is Correct explanation for Statement 1.
    B
    Statement 1 is True, Statement 2 is True, Statement 2 is NOT a correct explanation for Statement 1.
    C
    Statement 1 is True, Statement 2 is False.
    D
    Statement 1 is False, Statement 2 is True.

    • Similar Questions

    Explore conceptually related problems

    A and B are two isotopes having mass numbers 14 and 16 respectively. If the atomic number of A is 7, then how many neutrons will be there in B?

    The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are '2,4,10,12,14'. Find the remaining two observations:

    The mean and variance of 7 observations are 8 and 16 respectively . If five of the observations are 2,4,10,12,14 find the remaining two observations.

    A mixture of ""^(239)Pu and ""^(240)Pu has an activity of 6.0 xx 10^(9) dis/s. The half-life lives of isotopes are 2.44 xx 10^(4) and 6.58 xx 10^(3) years, respectively. Calculate the isotopic composition of this sample.

    Isotopes are the atoms of same element: they have same atomic number but different mass numbers. Isoto have different number of neutrons in their nucleus. If an element exists in two isotopes having atomic mas 'a' and 'b' in the ratio m:n, then average atomic mass will be ( m xx a + n xx b)/( m + n) Different isotopes of same element have same position in the periodic table. The elements which have sin isotope are called monoisotropic elements. Greater is the percentage composition of an isotope, more wiki be its abundance in nature. Atomic mass of boron is 10.81. It has two isotopes namely ""_(5) ^(11)B and ""_(5) ^(x) B with their relative abundance of 8 and 20% respectively. The value of x is

    Home
    Profile