How many atoms are there in 79.3 g of potassium dihydrogen phosphate (KH2PO4)?

Solutions

Expert Solution

Weight of postassium dihydrogen phosphate ( KH₂PO₄) = 79.3 grams

Molecular weight of potassium dihydrogen phosphate ( KH₂PO₄) =39+2(1)+31+16(4)=136 g/mole

Hence number of moles of potassium dihydrogen phosphate ( KH₂PO₄)

=Weight of potassium dihydrogen phosphate ( KH₂PO₄)/molecular weight of potassium dihydrogen phosphate (KH₂PO₄)

=79.3/136 =0.583 moles of KH₂PO₄

Now to calculate the number of atomes present in 79.3 grams of KH₂PO₄ we do the following.

We know that number of atomes in 1 moles = 6.023*10²³

There for number of atoms in 0.583 moles of KH₂PO₄=0.583*6.023*10²³=3.5119*10²³ atoms

Hence number of atoms in 79.3 g of KH₂PO₄ is 3.5119*10²³ atoms

queen_honey_blossom answered 1 year ago

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