Question

Volume of a mixture of 6.02 x 10^23 oxygen atoms and
3.01 x 10^23 hydrogen molecules at NTP is:​

Answer 1

2H2(g) + O2 ———> 2H2O

No. of moles of H2
$= \frac{6.02 \times {10}^{23} }{6.02 \times {10}^{23}} \\ = 1 \: mol$

No.of moles of O2
$= \frac{3.01\times {10}^{23}}{6.02 \times {10}^{23} } \\ 2 \: mol$

A/Q,
2 moles of H2 and 1 mole of O2 combined to from two moles of H2O.

=> 2 moles of H2 and 0.5 mole of O2 will combine to form one mole of H2O.

Therefore, 1 mole of H2O = 22. 7L

Hence, the volume of of H2 is 22. 7L.








HOPE IT'LL HELP.. :)

Answer 2

Answer:

22.4 litres

Explanation:

Number of Oxygen atoms = 6.022*10^23 atoms

So, number of Oxygen Molecules = 3.011*10^23 [since Oxygen is Diatomic, O2]

3.011*10^23 molecules is equal to 0.5 mol [By Calculation]

 

Number of Hydrogen molecules = 3.011*10^23 [Given]

3.011*10^23 molecules is equal to 0.5 mol [By Calculation]

 

When they are mixed together, then total number of moles is 1 mol [0.5 mol + 0.5 mol]

At NTP, 1 mole occupies 22.4 litres. Hence the correct option is (d)

 

Hope it helps...

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