Calculate the number of molecules of ammonia present in $5.6 d m^{3}$ of its volume.

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Solution

$V = 5.6 d m^{3} = 5.6 L$

Then ,
We know that,

$n = V / 22.4$
$n = 5.6 / 22.4$
$n = 56 / 224$
$n = 0.25$ mole.
And for the no of molecules,

We have a concept,

No. of molecules $= n$ $\times$ Avogadro number
No. of molecule $= 0.25 \times 6.022 \times 10^{23}$
No. of molecules $= 1.5 \times 10^{23}$

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Calculate the number of molecules of ammonia present in 5.6 dm3 of its volume.

V=5.6 dm3=5.6 LThen ,We know that,n=V/22.4n=5.6/22.4n=56/224n=0.25 mole.And for the no of molecules,We have a concept,No. of molecules =n × Avogadro numberNo. of molecule =0.25×6.022×1023No. of molecules =1.5×1023

Calculate the number of molecules of ammonia present in $5.6 d m^{3}$ of its volume.