Question

How many atoms are in 25.00 g of B.

1.393 x 10^24 atoms of B


1.333 x 10^23 atoms of B


2.16 x 10^26 atoms of B


2.17 x 10^24 atoms of B

Answer 1

`\boxed {\boxed {\sf 1.393 *10^(24) \ atoms \ B}}`

Step-by-step explanation:

1. Convert Grams to Moles

Use the molar mass (found on the Periodic Table) to convert from grams to moles.

• Boron (B): 10.81 g/mol

Use this value as a ratio.

`\frac {10.81 \ g \ B }{1 \ mol \ B}`

Multiply by the given number of grams.

`25.00 \ g \ B *\frac {10.81 \ g \ B }{1 \ mol \ B}`

Flip the ratio so the grams of boron cancel out.

`25.00 \ g \ B *\frac {1 \ mol \ B }{10.81 \ g \ B}`

`25.00 *\frac {1 \ mol \ B }{10.81 }`

`\frac {25.00 \ mol \ B }{10.81 }=2.312673451 \ mol \ B`

2. Convert Moles to Atoms

We use Avogadro's Number, 6.02*10²³: the number of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are atoms of boron.

`\frac {6.02*10^(23) \ atoms \ B} {1 \ mol \ B}`

Multiply by the number of moles we calculated.

`2.312673451 \ mol \ B *\frac {6.02*10^(23) \ atoms \ B} {1 \ mol \ B}`

The moles of boron cancel.

`2.312673451 *\frac {6.02*10^(23) \ atoms \ B} {1 }`

`2.312673451 *6.02*10^(23) \ atoms \ B} =1.39269195*10^(24) \ atoms \ B`

The original value of grams has 4 significant figures, so our answer should have the same. For the number we calculated, that is the thousandth place.

`1.392\underline69195*10^(24) \ atoms \ B`

The 6 tells us to round the 2 to a 3.

`1.393 *10^(24) \ atoms \ B`

25.00 grams of boron is equal to 1.393*10²⁴ atoms.