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Q.

1.0 Kg of Iron (Fe), having atomic mass equal to $56 g m o l^{- 1}$ contains :-

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a

1.075 \times 10^{25} a t o m s

b

6.93 \times 10^{23} a t o m s

c

6.93 \times 10^{21} a t o m s

d

2.88 \times 10^{24} a t o m s

answer is D.

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Detailed Solution

N o . o f a t o m s = n u m b e r o f m o l e s \times A v o g a d r o N o .

$N_{F e} = \frac{1000}{56} \times 6.022 \times 10^{23}$

$N_{F e} = 1.075 \times 10^{25} a t o m s$

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