question_answer 49)

  The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following elements contains the greatest number of atoms? (a) 4 g He                             (b) 46 g Na (c) 0.40 g Ca                        (d) 12 g He

Answer:

  (d) (a) Number of atoms of \[He=\frac{Mass}{Molar\,mass}\times 6.023\times {{10}^{23}}\] \[=\frac{4}{4}\times 6.023\times {{10}^{23}}\] \[=6.023\times {{10}^{23}}\] (b) Number of atoms of \[Na=\frac{46}{23}\times 6.023\times {{10}^{23}}\]                                                  \[=2\times 6.023\times {{10}^{23}}\] (c) Number of atoms of \[Ca=\frac{0.4}{40}\times 6.023\times {{10}^{23}}\] (d) Number of atoms of \[He=\frac{12}{4}\times 6.023\times {{10}^{23}}\] \[=3\times 6.023\times {{10}^{23}}\] (Greatest)