• # question_answer49)   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

(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)