Question

4.00 × 1023 molecules of H3PO4 have an approximate mass of

Answer 1

1 mole --------- 6.02 x 10²³ molecules
? mole --------- 4.00 x 10²³ molecules

moles = (4.00 x 10²³) x 1 / 6.02 x 10²³

moles = ( 4.00 x 10²³) / ( 6.02 x 10²³ )

= 0.6644 moles

Molar mass H₃PO₄ = 98.0 g/mol

mass = 0.6644 x 98.0 = 65.1112 g

hope this helps!

Answer 2

Answer : The approximate mass of
`H_3PO_4` is 65 grams.

Explanation :

First we have to calculate the moles of
`H_3PO_4`.

As,
`6.022* 10^(23)` molecules of
`H_3PO_4` present in 1 mole of
`H_3PO_4`

So,
`4.00* 10^(23)` molecules of
`H_3PO_4` present in
`(4.00* 10^(23))/(6.200* 10^(23))=0.664` mole of
`H_3PO_4`

Now we have to calculate the mass of
`H_3PO_4`.

`\text{Mass of }H_3PO_4=\text{Moles of }H_3PO_4* \text{Molar mass of }H_3PO_4`

Molar mass of
`H_3PO_4` = 98 g/mole

`\text{Mass of }H_3PO_4=0.664mole* 98g/mole=65.072g\approx 65g`

Therefore, the approximate mass of
`H_3PO_4` is 65 grams.