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Which of the following samples contains the smallest number of atoms

(A) 1gm of CO2  (B) 1gm of C8H18  (C) 1gm of C2H6. (D) 1gm of B4H10

Asked by djshah.jain4 10th February 2018, 10:02 AM
Answered by Expert
Answer:
First, find out the no. of atoms present in each case,
begin mathsize 22px style 1 right parenthesis space 1 space g m space o f space C O subscript 2 space equals fraction numerator 1 space m o l space C O subscript 2 over denominator 44 space g m space o f space C O subscript 2 end fraction space cross times fraction numerator 6.023 cross times 10 to the power of 23 a t o m s over denominator 1 space m o l space o f space C O subscript 2 end fraction space equals space 0.13 space cross times 10 to the power of 23 space space a t o m s
2 right parenthesis space 1 space g m space o f space C subscript 8 H subscript 18 space equals space fraction numerator 1 space m o l e space C subscript 8 H subscript 18 over denominator 114 space g m space o f space space C subscript 8 H subscript 18 space end fraction cross times fraction numerator 6.023 cross times 10 to the power of 23 a t o m s over denominator 1 space m o l space o f space C subscript 8 H subscript 18 end fraction space equals space 0.05 cross times 10 to the power of 23 space a t o m s
3 right parenthesis space 1 space g m space o f space C subscript 2 H subscript 6 space equals fraction numerator 1 space m o l e C subscript 2 H subscript 6 over denominator 30 space g m space o f space C subscript 2 H subscript 6 space end fraction cross times fraction numerator 6.023 cross times 10 to the power of 23 a t o m s over denominator 1 space m o l space o f space C subscript 2 H subscript 6 end fraction space equals space 0.2 cross times 10 to the power of 23 space a t o m s
4 right parenthesis space 1 space g m space o f space B subscript 4 H subscript 10 space equals space fraction numerator 1 space m o l e space B subscript 4 H subscript 10 over denominator 53.32 space g m space o f space B subscript 4 H subscript 10 space end fraction cross times fraction numerator 6.023 cross times 10 to the power of 23 a t o m s over denominator 1 space m o l space o f space B subscript 4 H subscript 10 end fraction space equals space 0.1 space cross times 10 to the power of 23 space a t o m s end style
 
 
from above values, we can conclude that option 2 1gm of  C8H18  has less no. of atoms
Answered by Expert 10th February 2018, 12:29 PM
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