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Some Basic Concepts Of Chemistry

Question
CBSEENCH11008028

Calculate the number of molecules present:

(i) in one drop of water having mass of 0.05 g

(ii) in 34.20 gram of cane sugar (C12H22O11).

Solution

(i) 1 mole of H2O = 18 g
                          = 6.023 x 1023 molecules
  Mass of 1 drop of water = 0.05 g
   Now 18 g of H2O contains 6.023 x 1023 molecules
              therefore 0.05 g of straight H subscript 2 straight O would contain
                                 equals space fraction numerator 6.023 space cross times space 10 to the power of 23 over denominator 18 end fraction cross times space 0.05 space equals space 1.673 space cross times space 10 to the power of 21 space molecules
equals space 1.673 space cross times space 10 to the power of 21 space molecules

(ii) Molecular mass of straight C subscript 2 straight H subscript 22 straight O subscript 11
  equals space 12 space cross times space 12 space plus 22 space cross times space 1 plus space 11 cross times 16
equals space 342 space amu
1 space mole space of space straight C subscript 12 straight H subscript 22 straight O subscript 11
space space space space space space space space space space space space space equals 342 space straight g
space space space space space space space space space space space space space equals space 6.023 space cross times space 10 to the power of 23 space molecules
Now space 342 space straight g space of space cane space sugar space contains
space space space space space space space space equals 6.023 space cross times space 10 to the power of 23 space molecules
therefore space space 34.2 space straight g space of space cane space sugar space would space contain
space space space space space space space space equals space fraction numerator 6.023 space cross times space 10 to the power of 23 over denominator 342 end fraction cross times space 34.2
space space space space space space space equals 6.023 space cross times space 10 to the power of 22 space molecules