• # question_answer 10 gms each of $C{{O}_{2}},\text{ }N{{H}_{3}}\text{ }and\text{ }{{O}_{2}}$ were taken in three separate flasks. What is the correct decreasing order of atoms? A) $C{{O}_{2}},\text{ }N{{H}_{3}},\text{ }{{O}_{2}}$  B) $N{{H}_{3}},\text{ }{{O}_{2}},\text{ }C{{O}_{2}}$C) ${{O}_{2}},\text{ }N{{H}_{3}},\text{ }C{{O}_{2}}$  D) $N{{H}_{3}},\text{ }C{{O}_{2}},\text{ }{{O}_{2}}$

Solution :

$Number\text{ }of\text{ }atoms\text{ }of\,\,C{{O}_{2}}=3\times \left( \frac{10}{44} \right)\times N$ $Number\text{ }of\text{ }atoms\text{ }of\,\,N{{H}_{3}}=4\times \left( \frac{10}{17} \right)\times N$$Number\text{ }of\text{ }atoms\text{ }of\,\,{{O}_{2}}=2\times \left( \frac{10}{32} \right)\times N$ Here N is Avogadro?s number. Hence the correct order is: $N{{H}_{3}},\,\,C{{O}_{2}},\,\,{{O}_{2}}$

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