Question

What is the mass (in grams) of 9.49 × 1024 molecules of methanol (CH3OH)?

Answer 1

`504.91gCH_(3)OH`

Step-by-step explanation:

To solve the question the following information is needed:

Avogadro´s number:
`6.022*10^(23)mol^(-1)`

Molar mass of methanol:
`32.04(g)/(mol)`

Then use the Avogadro´s number and the molar mass of the methanol to find the mass:

`9.49*10^(24)moleculesCH_(3)OH*(1molCH_(3)OH)/(6.022*10^(23)moleculesCH_(3)OH)*(32.04gCH_(3)OH)/(1molCH_(3)OH)=504.91gCH_(3)OH`

Answer 2

1 mol of any substance contains 6.022 x 10^23 units. These units could be atoms that make up an element or molecules that make up a compound.
1 mol of methanol contains 6.022 x10^23 molecules of methanol.
Mass of one mole of methanol is 32.04 g
6.022x10^23 molecules of methanol weighs 32.04 g
Therefore 9.49x10^24 molecules of methanol weigh - 32.04/(6.022x10^23) x (9.49 x10^24 ) = 504.9 g