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1.28    Which one of the following will have the largest number of atoms?
                (i) 1 g Au (s)
                (ii) 1 g Na (s)
                (iii) 1 g Li (s)
                (iv) 1 g of Cl2 (g)

Answers (2)

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Calculating and then comparing for each:

(i) 1 g of Au will contain:

= \frac{1}{197}\ mol= \frac{1}{197}\times6.022\times10^{23}\ atoms.

(ii) 1 g of Na will contain:

= \frac{1}{23}\ mol= \frac{1}{23}\times6.022\times10^{23}\ atoms. 

(iii) 1 g of Li will contain:

= \frac{1}{7}\ mol= \frac{1}{7}\times6.022\times10^{23}\ atoms.

(iv) 1 g of Cl2 will contain:

= \frac{1}{71}\ mol= \frac{1}{71}\times6.022\times10^{23}\ atoms.

Clearly, we can compare and say that the number of atoms in 1g of Li has the largest.

Posted by

Divya Prakash Singh

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Posted by

Shalini

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