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3 years ago

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

Chemistry

2 answers:

Furkat [3]3 years ago

8 0

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

Sloan [31]3 years ago

6 0

Answer: 500 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {avogadro's number}}=\frac{9.42\times10^{24}}{6.023\times 10^{23}}=15.6moles$

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

Thus 15.6 moles of $CH_3OH$ weigh= $\frac{32.04}{1}\times 15.6=500g$

Thus the mass of $9.42\times10^{24}$ molecules of methanol is 500 grams

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