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An unknown chlorohydrocarbon has 3.55% o...

An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in 1 g of chlorohydrocarbon are :
(Atomic wt. of Cl = 35.5 u,
Avogardo constant `=6.023xx10^(23) "mol"^(-1)`)

A

`6.023 xx 10^(9)`

B

`6.023 xx 10^(23)`

C

`6.023 xx 10^(21)`

D

`6.023 xx 10^(20)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given percentage of chlorine in an hydrocarbon = 3.55% i.e.,
100 g of chlorohydrocarbon has 3.55 g of chlorine.
1 gof chlorohydrocarbon will have `(355)/(100)= 0.0355 g` of chlorine.
Atomic wt. of `CI=35.5 g // mol`
Number of moles of `Cl =(0.0355 g)/(35.5 g // mol) = 0.001` mole
Number of atons of `CI=0.001" mole" xx 6.023 xx 10^(23) mol^(-1)`
`= 6.023 xx 10^(20)`
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Knowledge Check

  • An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in 1 g of chlorohydrocarbon are : (Atomic wt. of C1 = 35.5 u, Avogador constant = 6.023xx10^(23) mol^(-1))

    A
    `6.023xx10^(21)`
    B
    `6.023xx10^(23)`
    C
    `6.023xx10^(20)`
    D
    `6.023xx10^(9)`

  • Calculate the number of atoms present in 7.1 g of chlorine ?

    A
    `0.2 N_(A)`
    B
    `0.3N_(A)`
    C
    `0.4N_(A)`
    D
    `0.5 N_(A)`

  • The number of atoms in 0.1 mol of a triatomic gas is (N_A = 6.023 xx 10^(23) mol^(-1))

    A
    `6.023 xx 10^(22)`
    B
    `1.806 xx 10^(22)`
    C
    `3.600 xx 10^(23)`
    D
    `1.800 xx 10^(22)`

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