Question

at certain conditions of temperature and pressure 1 L of CH4 had contains 12 ×10^24 atoms. the volume occupied by SO3 gas at same conditions of temperature and pressure containing same number of atoms as that of CH4 sample is​

Answer 1

Answer:

The volume occupied by $SO_3$ gas at same conditions of temperature and pressure will be 1.3703 L.

Explanation:

The volume occupied by the methane gas at pressure and P and temperature T = 1 L

Number of atoms = $1\times 10^{24] atoms$

Number of molecules of methane be x

As we can see that 1 molecule of methane consists 5 atoms.

1 molecule methane = 5 atoms

$x \times 5=1\times 10^{24} atoms$

$x = 2.0\times 10^{23} molecules$

$1 mole = 6.022\times 10^{23} mol^{-1}$

$n=\frac{2.0\times 10^{23}}{6.022\times 10^{23} mol^{-1}}=0.3321 mol$

0.3321 moles of methane occupiers 1 L of volume at given conditions of temperature and pressure.

Number of molecules of $SO_3$ be y.

As we can see that 1 molecule of $SO_3$ consists 4 atoms.

1 molecule methane = 4 atoms

$y \times 4=1\times 10^{24] atoms$

$y = 2.5\times 10^{23} molecules$

$n=\frac{2.5\times 10^{23}}{6.022\times 10^{23} mol^{-1}}=0.4151 mol$

0.4151 moles of $SO_3$.

Assuming ,both the gases are behaving ideal to each other at given conditions of temperature and pressure.

1 L occupied by 0.3321 moles of methane.

1 mol will occupy =$\frac{1}{0.3321 mol} L$

Then 0.4151 moles of $SO_3$ will occupy:

$\frac{1}{0.3321 mol} L\times 0.4551 mol=1.3703 L$

The volume occupied by $SO_3$ gas at same conditions of temperature and pressure will be 1.3703 L.

Answer 2

Explanation:

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HOPE IT HELPS......

It may be easy than above ones

we can use avagadro law here