Question

What is the weight in grams of 3.36*10^23 molecules of copper sulfate

Answer 1

According to Avogadro's law, 1 mole ,i.e., 6.2 * 10^23 particles of every substance have mass equal to its molecular mass in grams. Since the molecular mass of CuSO4 is 80 grams. It means 6.2 * 10^23 particles of CuSO4 have a mass of 80 grams. Now u can calculate the mass of 3.36 * 10^23 particles by unitary method 3.36 * 10^23 *80/6.23 * 10^23 grams

Answer 2

The weight in grams of
`3.36*10^(23)` molecules of copper sulfate is 89.023 g.

Explanation:

Given : Molecules of copper sulfate
`3.36*10^(23)`

To find : What is the weight in grams of
`3.36*10^(23)` molecules of copper sulfate?

Solution :

Molar mass of copper sulfate (
`CuSO_4`)

Cu=63.5

S=32

O=16

`CuSO_4=63.5+32+4*16= 159.5 g`

We know, that 1 mole = Avogadro number

1 mole =
`6.02* 10^(23)` molecules

`6.02*10^(23)` molecule weigh = 159.5 g

1 molecule weigh =
`(159.5)/(6.02*10^(23))` g

So,
`3.36*10^(23)` molecule weigh =
`(159.5*3.36*10^23)/(6.02*10^(23))` g

`=(159.5*3.36)/(6.02)` g

`=(535.92)/(6.02)` g

`=89.023` g

Therefore, The weight in grams of
`3.36*10^(23)` molecules of copper sulfate is 89.023 g.