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

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Solution

Solution:
No. of moles= $\frac{v o l}{22.4}$
= $\frac{5.6}{22.4} = 0.25$ moles
No of molecules= $A v o g a d r o n o \times n o . o f m o l e s$
= $6.023 \times 10^{23} \times 0.25$
= $1.50575 \times 10^{23}$

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

Solution:No. of moles=vol22.4 =5.622.4=0.25 molesNo of molecules=Avogadro no×no. of moles =6.023×1023×0.25 =1.50575×1023

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

Solution:
No. of moles= $\frac{v o l}{22.4}$
= $\frac{5.6}{22.4} = 0.25$ moles
No of molecules= $A v o g a d r o n o \times n o . o f m o l e s$
= $6.023 \times 10^{23} \times 0.25$
= $1.50575 \times 10^{23}$