Question

What is the mass of 2.62 x 10 24 molecules of methanol (c3OH)

Answer 1

139.37 g of methanol is present in the given molecules.

Step-by-step explanation:

It is known that 1 mole of any compound contains avagadro's number of molecules. Since here the number of molecules of methanol is given as 2.62 ×
`10^(24)`, then the number of moles in them is

1 mole = 6.02 ×
`10^(23)` molecules

So 1 molecule =
`(1)/(6.022*10^(23))` moles

Thus, 2.62 ×
`10^(24)` molecules = 2.62 ×
`10^(24)`×
`(1)/(6.022*10^(23))` moles = 4.35 moles of methanol.

As it is known that 1 moles of any compound is equal to the molar mass of that compound. Here we know that 4.35 moles of methanol is present then, the mass is

Molar mass × Number of moles = Mass of methanol

32.04×4.35 = 139.37 g.

So 139.37 g of methanol is present in the given molecules.