The number of molecules in 5.65 g of ammonia is approximately x*10 raise to 23.what is the value of x

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Solution

Dear student

$N o . o f m o l e s = \frac{w e i g h t}{m o l e c u l a r m a s s} = \frac{5.65}{17} = 0.33 N o . o f m o l e c u l e s = n o . o f m o l e s \times 6.023 \times 10^{23} = 0.33 \times 6.023 \times 10^{23} ≅ 2 \times 10^{23}$

Hence value of x is 2

Regards

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The number of molecules in 5.65 g of ammonia is approximately x*10 raise to 23.what is the value of x

Dear student No.of moles =weightmolecular mass =5.6517 =0.33No.of molecules = no.of moles ×6.023×1023 = 0.33 ×6.023×1023 ≅2 ×1023 Hence value of x is 2 Regards

Dear student

$N o . o f m o l e s = \frac{w e i g h t}{m o l e c u l a r m a s s} = \frac{5.65}{17} = 0.33 N o . o f m o l e c u l e s = n o . o f m o l e s \times 6.023 \times 10^{23} = 0.33 \times 6.023 \times 10^{23} ≅ 2 \times 10^{23}$

Hence value of x is 2

Regards