what would be the mass of 1.98 x 10^28 molecules of h₂o?

Explanation:

Step1: Calculate the number of moles

Use Avogadro's number ($N_A = 6.022\times 10^{23}$ molecules/mol). The number of moles $n$ is given by $n=\frac{N}{N_A}$, where $N = 1.98\times 10^{28}$ molecules. $n=\frac{1.98\times 10^{28}\text{ molecules}}{6.022\times 10^{23}\text{ molecules/mol}}\approx 3.29\times 10^{4}\text{ mol}$

Step2: Calculate the molar - mass of water

The molar - mass of $H_2O$: The molar mass of $H$ is approximately $1.01$ g/mol and the molar mass of $O$ is approximately $16.00$ g/mol. So, $M_{H_2O}=2\times1.01\text{ g/mol}+ 16.00\text{ g/mol}=18.02\text{ g/mol}$

Step3: Calculate the mass of water

Use the formula $m = n\times M$. Substitute $n = 3.29\times 10^{4}\text{ mol}$ and $M = 18.02\text{ g/mol}$ into the formula. $m=(3.29\times 10^{4}\text{ mol})\times(18.02\text{ g/mol})\approx 5.93\times 10^{5}\text{ g}$

Answer:

$5.93\times 10^{5}\text{ g}$