Chemistry
Grade-9
Easy

Question

0.1g if carbon dioxide occupies a volume of 320cc at certain conditions. Under similar condition 0.2g of dioxide of element ‘X’ occupies 440cc. Calculate the atomic weight ‘X’.

  1. 34
  2. 39
  3. 32
  4. 38

The correct answer is: 32


    According to the ideal gas equation, PV = nRT. Here,
    V1 = 320 cc = 0.32 L
    m of CO2 = 12 + 16 x 2 = 12 + 32 = 44
    n1 = w/m = 0.1/44 gm = 0.0022
    V2 = 440 cc = 0.44 L
    V subscript 1 over n subscript 1 equals V subscript 2 over n subscript 2
fraction numerator 0.32 over denominator 0.0022 end fraction equals fraction numerator 0.44 over denominator n subscript 2 end fraction
n subscript 2 space equals space fraction numerator 0.44 space x space 0.0022 over denominator 0.32 end fraction equals 0.0030
    Now, n2 = w/m
    n subscript 2 equals fraction numerator 0.2 over denominator 32 plus x end fraction
0.0030 space equals space fraction numerator 0.2 over denominator 32 plus x end fraction
x space plus space 32 space equals space fraction numerator 0.2 over denominator 0.003 end fraction equals 66.67
x space equals space 66.67 space minus space 32 space equals space 34.67
    Hence, x = 34

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