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CBSE Class 11-science Answered

ans with solutions
Asked by anilsolanki2060 | 22 Feb, 2020, 10:12: AM
Expert Answer
Option (A) is correct.
 
Given:
 
Density of CO2 = 0.44 g/cm3 
 
We know,
 
1 mole of CO2 = 6.022 × 1023 molecules
 
6.022 × 1023 molecules = 44 gm of  CO2 
 
Mass of one molecules of  CO2 equals fraction numerator 44 over denominator 6.022 cross times 10 to the power of 23 end fraction
 
                                             equals 44 over straight N space
Density equals Mass over Volume
 
Volume
 
 
          equals Mass over Density

equals fraction numerator begin display style bevelled 44 over straight N end style over denominator 0.44 end fraction
 
 
 
We have,
 
For critical volume
 
    space space space straight V subscript straight c equals space 3 cross times 4 over 3 πr cubed

fraction numerator begin display style bevelled 44 over N end style over denominator 0.44 end fraction equals space 3 cross times 4 over 3 cross times πr cubed

space space space space space space straight r cubed space equals space fraction numerator 44 over denominator 0.44 cross times 4 straight pi cross times straight N end fraction

space space space space space space straight r cubed space equals 25 over πN
Answered by Varsha | 24 Feb, 2020, 11:20: AM

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