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The mole fraction of nitrogen, in a mixt...

The mole fraction of nitrogen, in a mixture containing 70 grams nitrogen, 120 grams of oxygen and 44 grams of carbon dioxide is

A

0.36

B

0.34

C

0.29

D

5

Text Solution

Verified by Experts

The correct Answer is:
B
Weight of nitrogen = 70 g, Weight of oxygen = 120 g and weight of carbon dioxide = 44 g.
Moles of `N_2 (n_1) = ("Weight")/("Molecular weight")= 70/28 =2.5`
Similarly, moles of `O_2 (n_2) = 120/32 = 3.75`
and moles of `CO_2 (n_3) = 44/44= 1`
Therefore mole fraction of nitrogen `(N_2)`
`= (n_1)/(n_1+n_2 + n_3) = (2.5)/(2.5+3.75 +1)= (2.5)/(7.25)=0.34`
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Knowledge Check

  • The mole fraction of oxygen in a mixture of 7g of nitrogen and 8g of oxygen is:

    A
    `(8)/(15)`
    B
    `0.5`
    C
    `0.25`
    D
    `1.0`

  • The mole fraction of nitrogen , in a mixture of a solution have a sum of ______.

    A
    `0.5`
    B
    `0.75`
    C
    `0.66`
    D
    `0.33`

  • One gram of oxygen is equal to

    A
    `10^(5) mu g`
    B
    `10^(6) mu g`
    C
    `10^(4) mu g`
    D
    `10^(3) mu g`

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