Question

What is the mass (in grams) of 9.42 × 1024 molecules of methanol (CH3OH)?

Answer 1

(9.42x10^24)/(6.02x10^23)=15.6478 mol methanol
15.6478mol(32.04g/1 mol)=501.3568g methanol

Answer 2

Answer: 500 grams

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 10^(23)` of particles.

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given molecules}}{\text {avogadro's number}}=(9.42*10^(24))/(6.023* 10^(23))=15.6moles`

1 mole of
`CH_3OH` weighs = 32.04 g/mol

Thus 15.6 moles of
`CH_3OH` weigh=
`(32.04)/(1)* 15.6=500g`

Thus the mass of
`9.42*10^(24)` molecules of methanol is 500 grams