Question

what would be the mass of 1.98x10^28 molecules of water

Answer 1

1st we need to calculate the number of moles that are present
$(1.98* 10^{28} atoms)( \frac{1 mol}{6.023*10^{23} atoms} ) = 32873.98 mol$ of water

2nd take the number of moles and multiply it by the Molecular Weight of water (18)
32873.98 * 18 = 591731.69 g = 591.7 kg

Answer 2

Answer: The mass of water would be $5.918\times 10^5g$

Explanation:

According to the mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules which determines the molar mass of that compound.

Mass of water for 1 mole is 18 grams.

So, $1.98\times 10^{28}$ number of molecules will have a mass of

$\frac{1.98\times 10^{28}}{6.022\times 10^{23}}\times 18g=5.918\times 10^5g$

Hence, the mass of water would be $5.918\times 10^5g$