What is the mass, in grams, of 9.89×1024 molecules of methanol (CH3OH)?

Answer 1

Did you mean 9.89 x 10²⁴? If so. this is a number greater than Avogadro's number which is 6.022 x 10²³ molecules per mole. An Avogadro's number of methanol molecules would have a mass of about 32 grams. So, as 9.89 x 10²⁴ is greater than Avogadro's number, the mass of this number of molecules would be greater than 32 g of methanol by a factor of 9.89 x 10²⁴ / 6.022 x 10²³ or about 16.4 times 32 g.