Question

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

Answer 1

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

Answer 2

Answer:

$504.91gCH_{3}OH$

Explanation:

To solve the question the following information is needed:

Avogadro´s number: $6.022*10^{23}mol^{-1}$

Molar mass of methanol: $32.04\frac{g}{mol}$

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

$9.49*10^{24}moleculesCH_{3}OH*\frac{1molCH_{3}OH}{6.022*10^{23}moleculesCH_{3}OH}*\frac{32.04gCH_{3}OH}{1molCH_{3}OH}=504.91gCH_{3}OH$