Question

How many silver atoms are present in a piece of jewellery weighing 10.78g? Ag=107.8
a.m.u.
ALL
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Answer 1

Answer:

10.7 g will contain 6.073 x 10²³ number of atoms.

Explanation:

It is given to us that :

a piece of jewellery weighs 10.78g which is also made of silver.

So, weight of silver, Ag = 107.8g

Hence, one mole of silver contains 107.8g = 107.8 a.m.u = 6.023 x 10²³ atoms

which means 107.8 g contains 6.023 x 10²³ atoms

so, 1 g will contain (6.023 x 10²³) / 107.8  = 5.587 x 10²¹ atoms

Therefore, 10.7 g will contain 5.587 x 10²¹ x 10.78 atoms = 6.073 x 10²³ number of atoms.

Hence, 10.7 g will contain 6.073 x 10²³ number of atoms.

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Answer 2

10.7 g will contain 6.073 x 10²³ number of atoms.

Explanation:

It is given to us that :

a piece of jewellery weighs 10.78g which is also made of silver.

So, weight of silver, Ag = 107.8g

Hence, one mole of silver contains 107.8g = 107.8 a.m.u = 6.023 x 10²³ atoms

which means 107.8 g contains 6.023 x 10²³ atoms

so, 1 g will contain (6.023 x 10²³) / 107.8  = 5.587 x 10²¹ atoms

Therefore, 10.7 g will contain 5.587 x 10²¹ x 10.78 atoms = 6.073 x 10²³ number of atoms.

Hence, 10.7 g will contain 6.073 x 10²³ number of atoms.

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