Question

What is the mass in grams of 9.06 X 10^24 molecules of methanol (CH3OH)?

Answer 1

the answer is 482.06g . . .

(9.06*10^24)(1mole/6.022*10^23)((MolarMass of CH3OH)/1mole) = 482.0606045g

Answer 2

Step-by-step explanation:

As we know that according to Avogadro, 1 mole of an atom contains
`6.022 * 10^(23)` atoms.

So, number of atoms given in the methanol are
`9.06 * 10^(24)`. Hence, calculate the number of moles present as follows.

No. of moles =
`(9.06 * 10^(24)atoms)/(6.022 * 10^(23) atoms/mol)`

= 15.04 mol

As molar mass of methanol is 32.04 g/mol. Hence, the mass of given methanol is as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

15.04 mol =
`(mass)/(32.04 g/mol)`

mass = 481.28 g

Thus, we can conclude that mass of
`9.06 * 10^(24)` molecules of methanol (
`CH_(3)OH`) is 481.28 g.