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If 3.01 xx 10^(28) molecules are removed...

If `3.01 xx 10^(28)` molecules are removed from 98 mg of `H_2SO_4,` then number of moles of `H_2SO_4` left are

A

`0.1 xx 10^(-2)` mol

B

`0.5 xx 10^(-3)` mol

C

`1.66 xx 10^(-3)` mol

D

`9.95 xx 10^(-2)` mol

Text Solution

Verified by Experts

The correct Answer is:
B
98 mg of
`H_2SO_4 = 6.02 xx 10^(23) xx 10^(-3) = 6.02 xx 10^(20)`
Number of `H_2SO_4`left
`= 6.02 xx 10^(29)- 3.01 xx 10^(20)`
`= 10^(20)(6.02 - 3.01) = 3.01 xx 10^(20)`
`because n = (N)/(N_A) = (3.01 xx 10^(20))/(6.02 xx 10^(23)) =0.5 xx 10^(-3)` mol
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Expess 9.8 g H_(2)SO_(4) in mole (Molecular mass of H_(2)SO_(4)=98 )

Knowledge Check

  • If 3.01xx10^(20) molecules are removed from 98 mg of H_2SO_4 , then number of moles of H_2SO_4 left are

    A
    `0.1xx10^(-3)`
    B
    `0.5xx10^(-3)` mol
    C
    `1.66xx10^(-3) mol
    D
    `9.95xx10^(-2)` mol.

  • If 3.01 xx 10^(20) molecules are removed from 98 mg of H_(2)SO_(4) , then number of moles of H_(2)SO_(4) left are

    A
    `0.5 xx 10^(-3)` mol
    B
    `0.1 xx 10^(-3)` mol
    C
    `9.95 xx 10^(-2)` mol
    D
    `1.66 xx 10^(-3)` mol

  • If 3.01xx10^20 molecules are removed from 98 mg of H_2SO_4 ,then number of moles of H_2SO_4 left are

    A
    `0.1 xx10^-3` mol
    B
    `0.5 Xx 10^-3 mol`
    C
    `1.66 x 10^-3 mol`
    D
    `9.95 xx 10^-2 mol`
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