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What is the mass (in grams) of 9.07 × 1024 molecules of methanol (CH3OH)?
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What is the mass (in grams) of 9.07 × 1024 molecules of methanol (CH3OH)?

User Robert Kolner
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1 Answer

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n=N/Na

=9.07x10^(24) / 6.02x10^(23)

= 15.1mol

m=n x M (molar mass of CH3OH is 32.05 g/Mol)

= 15.1 x 32.05

=484g

User Webaddicted
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