Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year
anam-khan

JEE Mains: Farmer's son from remote Maharashtra village aces exam

Latest Question