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  • Need explanation for: - Some basic concepts in chemistry - JEE Main-2

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;Avogadro constant=6.023\times 10^{23 } mol^{-1} )

  • Option 1)

    6.023\times 10^{20}

  • Option 2)

    6.023\times 10^{9}

  • Option 3)

    6.023\times 10^{21}

  • Option 4)

    6.023\times 10^{23}

 

Answers (2)

As we have learned

Number of Moles -

No of moles = given mass of substance/ molar mass of substance

-

 

 % of Cl = 3.55

weight of Cl = 1* 3.55/100= 0.0355 gm

moles of  Cl^{-} ion   = \frac{0.0355}{35.5}=1*10^{-3}

number of  Cl^{-} ion = 10^{-3}\times 6.022\times 10^{23} = 6.022 \times 10^{20 }

 

 

 

 


Option 1)

6.023\times 10^{20}

This is correct

Option 2)

6.023\times 10^{9}

This is incorrect

Option 3)

6.023\times 10^{21}

This is incorrect

Option 4)

6.023\times 10^{23}

This is incorrect

Posted by

Vakul

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