Question

The number of atoms present in 0.5 mole of nitrogen is same as the atoms in:
(a)12 g of C (b) 64 g of S
(c) 8 g of O (d) 48 g of mg

Answer 1

0.5 mole of Nitrogen

1 mole of $N=14$ gram

0.5 mole $N=0.5\times 14=7gram=\frac{1}{2}mole$

$=\frac{1}{2}\times 6.022\times {10}^{23}$atoms

$=3.011\times {10}^{23}atoms.$

(c) 8 gram O $=?$

16 gram of O $=1mole$

1 gram of O $=\frac{1}{16}mole$

8 gram of O $=\frac{1}{16}\times 8=\frac{1}{2}$

$=\frac{1}{2}\times 6.022\times {10}^{23}atoms=3.011\times {10}^{23}atoms.$

Same as in O,S mole of nitrogen.