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Serjik [45]
3 years ago
15

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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N=n\times N_A

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\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=33.35 \%

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\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=62.80\%

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