1
Question
how many molecules are present in
a) 9g of water
b) 17g of ammonium
how many molecules are present in
a) 9g of water
b) 17g of ammonium
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Solution
1 Mol of water = 18 g (2x1 for H +16 for O) = 6.023 x 1023
9 g of water = 0.5 moles = 6.023 x 1023 / 2 = 3.011 x 1023 molecules.
Ammonia has mol wt = N (14) + 3H =17
Thus 17g of ammonia will have 6.023 x 1023 molecules.
9 g of water = 0.5 moles = 6.023 x 1023 / 2 = 3.011 x 1023 molecules.
Ammonia has mol wt = N (14) + 3H =17
Thus 17g of ammonia will have 6.023 x 1023 molecules.
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