Data:
Molar Mass of CH3OH = 32.04 g/mol
\(9.66*10^{24}\:molecules\:of\:methanol=96.6*10^{23}\:molecules\:of\:methanol\)
Solving: According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
1 mol -------------------- 6.0210²³ molecules
y mol -------------------- 96.610²³ molecules
6.0210²³y = 96.610²³
\(y = \frac{96.6*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}\)
\(\boxed{y \approx 16.04\:mol}\)
Solving: Find the mass value now
32.04 g ----------------- 1 mol of CH3OH
x g ------------- 16.04 mol of CH3OH
\(x = 32.04*16.04\)
\(\boxed{\boxed{x = 513.92\:g}}\end{array}}\qquad\quad\checkmark\)
Answer:
The mass is 513.92 grams
