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The kinetic energy of a molecule of hydr...

The kinetic energy of a molecule of hydrogen at `0^(@) C` is `5.64 xx 10^(-21) J`. Calculate Avogadro's number. Take ` R = 8.31 J "mole"^(-1) K^(-1)`

A

`6.5xx10^(23)`

B

`6.43xx10^(23)`

C

`6.304xx10^(23)`

D

`6.034xx10^(23)`

Text Solution

Verified by Experts

The correct Answer is:
D
`((1)/(2)mv^(2))N_(A)=(3)/(2)RT`
`N_(A)=((3)/(2)xx8.31xx273)/(5.64xx10^(-21))N_(A)=6.034xx10^(-23)`
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The kinetic energy of a molecule of hydrogen at 0^(@) is 5.64 xx 10^(-21) J . Calculate Avogadro's number. Take R = 8.31 xx J "mole"^(-1) K^(-1)

The kinetic energy of a molecule of oxygen at 0^(@)C is 5.64 xx 10^(-21) J . Calculate Avogadro's number. Given R = 8.31 J "mol^(-1)K^(-1) .

Knowledge Check

  • The kinetic energy of two moles of N_2 at 27^(@)C is (R= 8.314 J K^(-1) "mol"^(-1) )

    A
    5491.6 J
    B
    6491.6 J
    C
    7482.6J
    D
    8882.4 J

  • The kinetic energy of two moles of N_(2) at 27^(@) C is (R = 8.314 J K^(-1) mol^(-1))

    A
    `5491.6 J`
    B
    `6491.6 J`
    C
    `7482.6 J`
    D
    `8880.4 J`

  • The average kinetic energy of gas molecule at 27^(@)C is 6.21xx10^(-21) J. Its average kinetic energy at 127^(@)C will be

    A
    `12.2xx10^(-21)J`
    B
    `8.28xx10^(-21)J`
    C
    `10.35xx10^(-21)J`
    D
    `11.35xx10^(-21)J`

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    Kinetic energy of oxygen molecule at 0^(@) C is 9.4 xx 10^(-21) j . Calculate Avogadro's number, when R = 8.31 J "mole"^(-1) K^(-1) .

    Given that the mass of a molecule of hydrogen is 3.34xx10^(-27) kg. Calculate Avogadro's number.

    The average kinetic energy of a gas molecule at 0^(@)C is 5.621xx10^(-21) J . Calculate Boltzmann costant. Also calculate the number of molecules present in one mole of the gas.

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    Calculate kinetic energy of a gram molecule of oxygen at 127^(@)C . Value of Boltzmann constant = 1.381 xx 10^(-23) J K^(-1) . Avogardro's number = 6.022 xx 10^(23) per gm-mole.

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