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Number of atoms present in 4.25g of 'NH3...

Number of atoms present in 4.25g of 'NH_3' is

A

`6.02xx10^(23)`

B

`4xx6.02xx10^(23)`

C

`1.7xx10^(24)`

D

`4.5xx6.02xx10^(23)`

Text Solution

Verified by Experts

The correct Answer is:
A
Number of molecules of `NH_(3)=(w)/(M)xx6.02xx10^(23)`
`=(4.25)/(17)xx6.02xx10^(23)" "` Number of atoms `=4xx(4.25)/(17)xx6.02xx10^(23)=6.02xx10^(23)`.
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Knowledge Check

  • Number of atoms in 4.25 g of NH_(3) is:

    A
    `6.02 xx 10 ^(23)`
    B
    `4 xx 6. 02 xx 10 ^(23)`
    C
    `1.7 xx 10 ^(24)`
    D
    `4.5 xx 6. 02 xx 10 ^(23)`

  • The number of atoms present in 16 g of oxygen is:

    A
    `6.02 xx 10^(11.5)`
    B
    `3.01 xx 10^(23)`
    C
    `3.01 xx 10^(11.5)`
    D
    `6.02 xx 10^(23)`

  • Number of atoms of oxygen present in 10.6g of Na_(2)CO_(3) will be

    A
    `6.02xx10^(22)`
    B
    `12.04xx10^(22)`
    C
    `1.806xx10^(23)`
    D
    `31.80xx10^(28)`

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