If one mole of aluminium contains 6 x 10²³ atoms of aluminium, how many atoms are contained in 0.9g of aluminium?
(Al = 27)

asuquoe528

4 years ago

If 27g of Al contain 6.02 x 10²³
0.9g of Al will contain:
0.9 x 6.02 x10²³/27
=2.0 x 10²². Copy that

Desfy

1 year ago

Here's how to solve this problem:

Step 1: Find the number of moles in 0.9g of aluminium

One mole of aluminium has a mass of 27g.
To find the number of moles in 0.9g, divide the mass by the molar mass: 0.9g / 27g/mol = 0.0333 moles
Step 2: Find the number of atoms in 0.0333 moles of aluminium

One mole of aluminium contains 6 x 10^23 atoms.
To find the number of atoms in 0.0333 moles, multiply the number of moles by Avogadro's number: 0.0333 moles * 6 x 10^23 atoms/mol = 2.0 x 10^22 atoms
Answer:

The answer is C. 2.0 x 10^22 atoms.

Anela

1 year ago

m=0.9
mm=27
n=1
Avogadro's constant=6.02×10^23


Therefore;n=m/mm
=0.9/27
=0.0333


1mol. =. 6.02×10^23
0.0333 mol = x
CROSS MULTIPLY

=6.02×10^23×0.0333/1

=2×10^22....ans

Uyihosa

6 months ago

I was sure my answer was correct, until I saw this explanation

Pabloxto

10 years ago

THIS WRONG!

27 - 6.0*10^23

0.9 - x

x=0.9*6*10^23/27

Pabloxto

10 years ago

On a second thought you're correct

BlackIce

1 year ago

wtf is that explanation 🤧🤧

Amuchie

10 years ago

Gud question