What mass of Ca will contain the same number of atoms as are present in 3.2 g of S?

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Solution

We know that,

Weigh of 1 mole of an element = Atomic mass of the element

Number of atoms in 1 mole of an element = 6.023 x

So, 32 g of S contains 6.023x atoms.

Therefore, 3.2 g of S will contain $equals fraction numerator 6.023 cross times 10 to the power of 23 cross times 3.2 over denominator 32 end fraction$ = 6.023xatoms of S.

Now, the weight of 6.023x atoms of Ca is 40 g.

Thus, the weight of 6.023x atoms of Ca will be $fraction numerator 6.023 cross times 10 to the power of 22 cross times 40 over denominator 6.023 cross times 10 to the power of 23 end fraction$ = 4 g.

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