Verified Answer :

Kalahira

You first calculate the formula mass of methanol (CH3OH). This is, add the mass of each element in the formula. C: 12 g/mol; H: 4 1 g/mol = 4 g/mol; O: 16 g/mol => formula mass = 12 g/mol + 4g/mol + 16 g/mol = 32 g/mol. Second, calculate the number of moles in 9,02 10^24 molecules, using Avogadros number: number of moles = 9.02 10^24 molecules / 6.02 10^23 molecules/mol = 15 moles. Third, the mass in grams of the compound equals the number of moles times the formula mass = 15 moles * 32 g/mol = 480 g. Answer: 480 grams.

Answer:

479.36 grams is the mass (in grams) of \(9.02\times 10^{24}\)  molecules of methanol .

Explanation:

\(n=\frac{m}{M}\)

\(N=n\times N_A\)

n = Moles of compound

m = mass of the compound

M = molar mass of the compound

N = Number of molecules or atoms of compounds

\(N_A=6.022\times 10^{23} mol^{-1}\)

Given :

Number of methanol molecules = N

\(N=9.02\times 10^{24}\) molecules

n = ?

\(N=n\times N_A\)

\(n=\frac{N}{N_A}=\frac{9.02\times 10^{24}}{6.022\times 10^{23} mol^{-1}}=14.98 mol\)

Molar mass of methanol = M = 32 g/mol

m = ?

\(n=\frac{m}{M}\)

\(m=n\times M=14.98 mol\times 32 g/mol=479.36 g\)

479.36 grams is the mass (in grams) of \(9.02\times 10^{24}\)  molecules of methanol .

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