Question

Calculate the no. of molecules and no. of atoms present in 11.2 l of oxygen at N. T. P.​

Answer 1

Answer:

No.of molecules=(6.022×10^23)/2=3.011×10^23

No.of atoms = 3.011×10^23=6.022×10^23

Answer 2

Q. Calculate the no. of molecules and no. of atoms present in 11.2 l of oxygen at N. T. P.

Ans:-

$no.of \: molecules = 3.01 \times {10}^{23}$

$no. \: of \: atoms = 6.02 \times {10}^{23}$

Explanation:-

$we \: know \: that \: 1 \: mole \: of \: oxygen \: at \: n.t.p. \: ocuppies \: 22.4l$

$= > 11.2l \: of \: oxygen \: at \: n.t.p. = \frac{1}{22.4} \times 11.2$

$= 0.5 \: mole$

$now \: 1 \: mole \: of \: oxygen \: contain \: 6.022 \times {10}^{23} \: molecules$

$0.5mole \: of \: oxygen \: contains \: = 6.022 \times {10}^{23} \times 0.5$

$= 3.01 \times {10}^{23} molecules$

$1molecule \: of \: oxygen \: = 2 \times 3.01 \times {10}^{23}$

$= 6.02 \times {10}^{23} atoms$